What do you call a linked list where every node has a pointer to the successor and predecessor?

Complex Linked List Implementation

The implementation we have learnt till now is Singly Linked List as it contains only link to a single successor. But there are also other complex linked list implementation in data structure.

Complex linked list implementation consists of three important implementations, i.e. the circularly linked list, the doubly linked list, and the multilinked list.

We wil discuss each of them now. Let’s learn about the first type, i.e. Circularly Linked List.

Circularly Linked List

In a circularly linked list, the points of the last node is linked to the first node of the list. They are mainly used in such list that allows access to the middle nodes of the list without starting at the beginning.

Doubly-linked list:

Linked List Types in Data Structure

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Inorder Successor in Binary Search Tree

In Binary Tree, Inorder successor of a node is the next node in Inorder traversal of the Binary Tree. Inorder Successor is NULL for the last node in Inorder traversal.

In Binary Search Tree, Inorder Successor of an input node can also be defined as the node with the smallest key greater than the key of the input node. So, it is sometimes important to find next node in sorted order.

In the above diagram, inorder successor of 8 is 10, inorder successor of 10 is 12 and inorder successor of 14 is 20.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1 (Uses Parent Pointer)
In this method, we assume that every node has a parent pointer.
The Algorithm is divided into two cases on the basis of the right subtree of the input node being empty or not.



Input: node, root // node is the node whose Inorder successor is needed.
Output: succ // succ is Inorder successor of node.

  1. If right subtree of node is not NULL, then succ lies in right subtree. Do the following.
    Go to right subtree and return the node with minimum key value in the right subtree.
  2. If right subtree of node is NULL, then succ is one of the ancestors. Do the following.
    Travel up using the parent pointer until you see a node which is left child of its parent. The parent of such a node is the succ.

Implementation:
Note that the function to find InOrder Successor is highlighted (with gray background) in below code.

C++




#include <iostream>
using namespace std;
/* A binary tree node has data,
the pointer to left child
and a pointer to right child */
struct node {
int data;
struct node* left;
struct node* right;
struct node* parent;
};
struct node* minValue(struct node* node);
struct node* inOrderSuccessor(
struct node* root,
struct node* n)
{
// step 1 of the above algorithm
if (n->right != NULL)
return minValue(n->right);
// step 2 of the above algorithm
struct node* p = n->parent;
while (p != NULL && n == p->right) {
n = p;
p = p->parent;
}
return p;
}
/* Given a non-empty binary search tree,
return the minimum data
value found in that tree. Note that
the entire tree does not need
to be searched. */
struct node* minValue(struct node* node)
{
struct node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL) {
current = current->left;
}
return current;
}
/* Helper function that allocates a new
node with the given data and
NULL left and right pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(
struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
node->parent = NULL;
return (node);
}
/* Give a binary search tree and
a number, inserts a new node with
the given number in the correct
place in the tree. Returns the new
root pointer which the caller should
then use (the standard trick to
avoid using reference parameters). */
struct node* insert(struct node* node,
int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return (newNode(data));
else {
struct node* temp;
/* 2. Otherwise, recur down the tree */
if (data <= node->data) {
temp = insert(node->left, data);
node->left = temp;
temp->parent = node;
}
else {
temp = insert(node->right, data);
node->right = temp;
temp->parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
/* Driver program to test above functions*/
int main()
{
struct node *root = NULL, *temp, *succ, *min;
// creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root->left->right->right;
succ = inOrderSuccessor(root, temp);
if (succ != NULL)
cout << "\n Inorder Successor of " << temp->data<< " is "<< succ->data;
else
cout <<"\n Inorder Successor doesn't exit";
getchar();
return 0;
}
// this code is contributed by shivanisinghss2110
C




#include <stdio.h>
#include <stdlib.h>
/* A binary tree node has data,
the pointer to left child
and a pointer to right child */
struct node {
int data;
struct node* left;
struct node* right;
struct node* parent;
};
struct node* minValue(struct node* node);
struct node* inOrderSuccessor(
struct node* root,
struct node* n)
{
// step 1 of the above algorithm
if (n->right != NULL)
return minValue(n->right);
// step 2 of the above algorithm
struct node* p = n->parent;
while (p != NULL && n == p->right) {
n = p;
p = p->parent;
}
return p;
}
/* Given a non-empty binary search tree,
return the minimum data
value found in that tree. Note that
the entire tree does not need
to be searched. */
struct node* minValue(struct node* node)
{
struct node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL) {
current = current->left;
}
return current;
}
/* Helper function that allocates a new
node with the given data and
NULL left and right pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(
struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
node->parent = NULL;
return (node);
}
/* Give a binary search tree and
a number, inserts a new node with
the given number in the correct
place in the tree. Returns the new
root pointer which the caller should
then use (the standard trick to
avoid using reference parameters). */
struct node* insert(struct node* node,
int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return (newNode(data));
else {
struct node* temp;
/* 2. Otherwise, recur down the tree */
if (data <= node->data) {
temp = insert(node->left, data);
node->left = temp;
temp->parent = node;
}
else {
temp = insert(node->right, data);
node->right = temp;
temp->parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
/* Driver program to test above functions*/
int main()
{
struct node *root = NULL, *temp, *succ, *min;
// creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root->left->right->right;
succ = inOrderSuccessor(root, temp);
if (succ != NULL)
printf(
"\n Inorder Successor of %d is %d ",
temp->data, succ->data);
else
printf("\n Inorder Successor doesn't exit");
getchar();
return 0;
}
Java




// Java program to find minimum
// value node in Binary Search Tree
// A binary tree node
class Node {
int data;
Node left, right, parent;
Node(int d)
{
data = d;
left = right = parent = null;
}
}
class BinaryTree {
static Node head;
/* Given a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
Node insert(Node node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null) {
return (new Node(data));
}
else {
Node temp = null;
/* 2. Otherwise, recur down the tree */
if (data <= node.data) {
temp = insert(node.left, data);
node.left = temp;
temp.parent = node;
}
else {
temp = insert(node.right, data);
node.right = temp;
temp.parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
Node inOrderSuccessor(Node root, Node n)
{
// step 1 of the above algorithm
if (n.right != null) {
return minValue(n.right);
}
// step 2 of the above algorithm
Node p = n.parent;
while (p != null && n == p.right) {
n = p;
p = p.parent;
}
return p;
}
/* Given a non-empty binary search
tree, return the minimum data
value found in that tree. Note that
the entire tree does not need
to be searched. */
Node minValue(Node node)
{
Node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null) {
current = current.left;
}
return current;
}
// Driver program to test above functions
public static void main(String[] args)
{
BinaryTree tree = new BinaryTree();
Node root = null, temp = null, suc = null, min = null;
root = tree.insert(root, 20);
root = tree.insert(root, 8);
root = tree.insert(root, 22);
root = tree.insert(root, 4);
root = tree.insert(root, 12);
root = tree.insert(root, 10);
root = tree.insert(root, 14);
temp = root.left.right.right;
suc = tree.inOrderSuccessor(root, temp);
if (suc != null) {
System.out.println(
"Inorder successor of "
+ temp.data + " is " + suc.data);
}
else {
System.out.println(
"Inorder successor does not exist");
}
}
}
// This code has been contributed by Mayank Jaiswal
Python3




# Python program to find the inorder successor in a BST
# A binary tree node
class Node:
# Constructor to create a new node
def __init__(self, key):
self.data = key
self.left = None
self.right = None
def inOrderSuccessor(n):
# Step 1 of the above algorithm
if n.right is not None:
return minValue(n.right)
# Step 2 of the above algorithm
p = n.parent
while( p is not None):
if n != p.right :
break
n = p
p = p.parent
return p
# Given a non-empty binary search tree, return the
# minimum data value found in that tree. Note that the
# entire tree doesn't need to be searched
def minValue(node):
current = node
# loop down to find the leftmost leaf
while(current is not None):
if current.left is None:
break
current = current.left
return current
# Given a binary search tree and a number, inserts a
# new node with the given number in the correct place
# in the tree. Returns the new root pointer which the
# caller should then use( the standard trick to avoid
# using reference parameters)
def insert( node, data):
# 1) If tree is empty then return a new singly node
if node is None:
return Node(data)
else:
# 2) Otherwise, recur down the tree
if data <= node.data:
temp = insert(node.left, data)
node.left = temp
temp.parent = node
else:
temp = insert(node.right, data)
node.right = temp
temp.parent = node
# return the unchanged node pointer
return node
# Driver program to test above function
root = None
# Creating the tree given in the above diagram
root = insert(root, 20)
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root.left.right.right
succ = inOrderSuccessor(temp)
if succ is not None:
print ("\nInorder Successor of % d is % d "%(temp.data, succ.data))
else:
print ("\nInorder Successor doesn't exist")
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
C#




// C# program to find minimum
// value node in Binary Search Tree
using System;
// A binary tree node
public
class Node
{
public
int data;
public
Node left, right, parent;
public
Node(int d)
{
data = d;
left = right = parent = null;
}
}
public class BinaryTree
{
static Node head;
/* Given a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
Node insert(Node node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null)
{
return (new Node(data));
}
else
{
Node temp = null;
/* 2. Otherwise, recur down the tree */
if (data <= node.data)
{
temp = insert(node.left, data);
node.left = temp;
temp.parent = node;
}
else
{
temp = insert(node.right, data);
node.right = temp;
temp.parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
Node inOrderSuccessor(Node root, Node n)
{
// step 1 of the above algorithm
if (n.right != null)
{
return minValue(n.right);
}
// step 2 of the above algorithm
Node p = n.parent;
while (p != null && n == p.right)
{
n = p;
p = p.parent;
}
return p;
}
/* Given a non-empty binary search
tree, return the minimum data
value found in that tree. Note that
the entire tree does not need
to be searched. */
Node minValue(Node node)
{
Node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null)
{
current = current.left;
}
return current;
}
// Driver program to test above functions
public static void Main(String[] args)
{
BinaryTree tree = new BinaryTree();
Node root = null, temp = null, suc = null, min = null;
root = tree.insert(root, 20);
root = tree.insert(root, 8);
root = tree.insert(root, 22);
root = tree.insert(root, 4);
root = tree.insert(root, 12);
root = tree.insert(root, 10);
root = tree.insert(root, 14);
temp = root.left.right.right;
suc = tree.inOrderSuccessor(root, temp);
if (suc != null) {
Console.WriteLine(
"Inorder successor of "
+ temp.data + " is " + suc.data);
}
else {
Console.WriteLine(
"Inorder successor does not exist");
}
}
}
// This code is contributed by aashish2995
Javascript




<script>
// JavaScript program to find minimum
// value node in Binary Search Tree
// A binary tree node
class Node {
constructor(val) {
this.data = val;
this.left = null;
this.right = null;
this.parent = null;
}
}
var head;
/*
* Given a binary search tree and a number,
inserts a new node with the given
* number in the correct place in the tree.
Returns the new root pointer which
* the caller should then use
(the standard trick to afunction using reference
* parameters).
*/
function insert(node , data) {
/*
* 1. If the tree is empty,
return a new, single node
*/
if (node == null) {
return (new Node(data));
} else {
var temp = null;
/* 2. Otherwise, recur down the tree */
if (data <= node.data) {
temp = insert(node.left, data);
node.left = temp;
temp.parent = node;
} else {
temp = insert(node.right, data);
node.right = temp;
temp.parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
function inOrderSuccessor(root, n) {
// step 1 of the above algorithm
if (n.right != null) {
return minValue(n.right);
}
// step 2 of the above algorithm
var p = n.parent;
while (p != null && n == p.right) {
n = p;
p = p.parent;
}
return p;
}
/*
* Given a non-empty binary search tree,
return the minimum data value found in
* that tree. Note that the entire tree
does not need to be searched.
*/
function minValue(node) {
var current = node;
/* loop down to find the leftmost leaf */
while (current.left != null) {
current = current.left;
}
return current;
}
// Driver program to test above functions
var root = null, temp = null,
suc = null, min = null;
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root.left.right.right;
suc = inOrderSuccessor(root, temp);
if (suc != null) {
document.write("Inorder successor of " +
temp.data + " is " + suc.data);
} else {
document.write(
"Inorder successor does not exist"
);
}
// This code contributed by gauravrajput1
</script>
Output Inorder Successor of 14 is 20

Complexity Analysis:

  • Time Complexity: O(h), where h is the height of the tree.
    As in the second case(suppose skewed tree) we have to travel all the way towards the root.
  • Auxiliary Space: O(1).
    Due to no use of any data structure for storing values.

Method 2 (Search from root)
Parent pointer is NOT needed in this algorithm. The Algorithm is divided into two cases on the basis of right subtree of the input node being empty or not.

Input: node, root // node is the node whose Inorder successor is needed.

Output: succ // succ is Inorder successor of node.

  1. If right subtree of node is not NULL, then succ lies in right subtree. Do the following.
    Go to right subtree and return the node with minimum key value in the right subtree.
  2. If right subtree of node is NULL, then start from the root and use search-like technique. Do the following.
    Travel down the tree, if a node’s data is greater than root’s data then go right side, otherwise, go to left side.

Below is the implementation of the above approach:

C++




// C++ program for above approach
#include <iostream>
using namespace std;
/* A binary tree node has data,
the pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
struct node* parent;
};
struct node* minValue(struct node* node);
struct node* inOrderSuccessor(struct node* root,
struct node* n)
{
// Step 1 of the above algorithm
if (n->right != NULL)
return minValue(n->right);
struct node* succ = NULL;
// Start from root and search for
// successor down the tree
while (root != NULL)
{
if (n->data < root->data)
{
succ = root;
root = root->left;
}
else if (n->data > root->data)
root = root->right;
else
break;
}
return succ;
}
// Given a non-empty binary search tree,
// return the minimum data value found
// in that tree. Note that the entire
// tree does not need to be searched.
struct node* minValue(struct node* node)
{
struct node* current = node;
// Loop down to find the leftmost leaf
while (current->left != NULL)
{
current = current->left;
}
return current;
}
// Helper function that allocates a new
// node with the given data and NULL left
// and right pointers.
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
node->parent = NULL;
return (node);
}
// Give a binary search tree and a
// number, inserts a new node with
// the given number in the correct
// place in the tree. Returns the new
// root pointer which the caller should
// then use (the standard trick to
// avoid using reference parameters).
struct node* insert(struct node* node,
int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return (newNode(data));
else
{
struct node* temp;
/* 2. Otherwise, recur down the tree */
if (data <= node->data)
{
temp = insert(node->left, data);
node->left = temp;
temp->parent = node;
}
else
{
temp = insert(node->right, data);
node->right = temp;
temp->parent = node;
}
/* Return the (unchanged) node pointer */
return node;
}
}
// Driver code
int main()
{
struct node *root = NULL, *temp, *succ, *min;
// Creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root->left->right->right;
// Function Call
succ = inOrderSuccessor(root, temp);
if (succ != NULL)
cout << "\n Inorder Successor of "
<< temp->data << " is "<< succ->data;
else
cout <<"\n Inorder Successor doesn't exit";
getchar();
return 0;
}
// This code is contributed by shivanisinghss2110
C




// C program for above approach
#include <stdio.h>
#include <stdlib.h>
/* A binary tree node has data,
the pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
struct node* parent;
};
struct node* minValue(struct node* node);
struct node* inOrderSuccessor(
struct node* root,
struct node* n)
{
// step 1 of the above algorithm
if (n->right != NULL)
return minValue(n->right);
struct node* succ = NULL;
// Start from root and search for
// successor down the tree
while (root != NULL)
{
if (n->data < root->data)
{
succ = root;
root = root->left;
}
else if (n->data > root->data)
root = root->right;
else
break;
}
return succ;
}
/* Given a non-empty binary search tree,
return the minimum data
value found in that tree. Note that
the entire tree does not need
to be searched. */
struct node* minValue(struct node* node)
{
struct node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL)
{
current = current->left;
}
return current;
}
/* Helper function that allocates a new
node with the given data and
NULL left and right pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(
struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
node->parent = NULL;
return (node);
}
/* Give a binary search tree and
a number, inserts a new node with
the given number in the correct
place in the tree. Returns the new
root pointer which the caller should
then use (the standard trick to
avoid using reference parameters). */
struct node* insert(struct node* node,
int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return (newNode(data));
else
{
struct node* temp;
/* 2. Otherwise, recur down the tree */
if (data <= node->data)
{
temp = insert(node->left, data);
node->left = temp;
temp->parent = node;
}
else
{
temp = insert(node->right, data);
node->right = temp;
temp->parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
/* Driver program to test above functions*/
int main()
{
struct node *root = NULL, *temp, *succ, *min;
// creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root->left->right->right;
// Function Call
succ = inOrderSuccessor(root, temp);
if (succ != NULL)
printf(
"\n Inorder Successor of %d is %d ",
temp->data, succ->data);
else
printf("\n Inorder Successor doesn't exit");
getchar();
return 0;
}
// Thanks to R.Srinivasan for suggesting this method.
Java




// Java program for above approach
class GFG
{
/* A binary tree node has data,
the pointer to left child
and a pointer to right child */
static class node
{
int data;
node left;
node right;
node parent;
};
static node inOrderSuccessor(
node root,
node n)
{
// step 1 of the above algorithm
if (n.right != null)
return minValue(n.right);
node succ = null;
// Start from root and search for
// successor down the tree
while (root != null)
{
if (n.data < root.data)
{
succ = root;
root = root.left;
}
else if (n.data > root.data)
root = root.right;
else
break;
}
return succ;
}
/* Given a non-empty binary search tree,
return the minimum data
value found in that tree. Note that
the entire tree does not need
to be searched. */
static node minValue(node node)
{
node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null)
{
current = current.left;
}
return current;
}
/* Helper function that allocates a new
node with the given data and
null left and right pointers. */
static node newNode(int data)
{
node node = new node();
node.data = data;
node.left = null;
node.right = null;
node.parent = null;
return (node);
}
/* Give a binary search tree and
a number, inserts a new node with
the given number in the correct
place in the tree. Returns the new
root pointer which the caller should
then use (the standard trick to
astatic void using reference parameters). */
static node insert(node node,
int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null)
return (newNode(data));
else
{
node temp;
/* 2. Otherwise, recur down the tree */
if (data <= node.data)
{
temp = insert(node.left, data);
node.left = temp;
temp.parent = node;
}
else
{
temp = insert(node.right, data);
node.right = temp;
temp.parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
/* Driver program to test above functions*/
public static void main(String[] args)
{
node root = null, temp, succ, min;
// creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root.left.right.right;
// Function Call
succ = inOrderSuccessor(root, temp);
if (succ != null)
System.out.printf(
"\n Inorder Successor of %d is %d ",
temp.data, succ.data);
else
System.out.printf("\n Inorder Successor doesn't exit");
}
}
// This code is contributed by gauravrajput1
Python3




# Python program to find
# the inorder successor in a BST
# A binary tree node
class Node:
# Constructor to create a new node
def __init__(self, key):
self.data = key
self.left = None
self.right = None
def inOrderSuccessor(root, n):
# Step 1 of the above algorithm
if n.right is not None:
return minValue(n.right)
# Step 2 of the above algorithm
succ=Node(None)
while( root):
if(root.data<n.data):
root=root.right
elif(root.data>n.data):
succ=root
root=root.left
else:
break
return succ
# Given a non-empty binary search tree,
# return the minimum data value
# found in that tree. Note that the
# entire tree doesn't need to be searched
def minValue(node):
current = node
# loop down to find the leftmost leaf
while(current is not None):
if current.left is None:
break
current = current.left
return current
# Given a binary search tree
# and a number, inserts a
# new node with the given
# number in the correct place
# in the tree. Returns the
# new root pointer which the
# caller should then use
# (the standard trick to avoid
# using reference parameters)
def insert( node, data):
# 1) If tree is empty
# then return a new singly node
if node is None:
return Node(data)
else:
# 2) Otherwise, recur down the tree
if data <= node.data:
temp = insert(node.left, data)
node.left = temp
temp.parent = node
else:
temp = insert(node.right, data)
node.right = temp
temp.parent = node
# return the unchanged node pointer
return node
# Driver program to test above function
if __name__ == "__main__":
root = None
# Creating the tree given in the above diagram
root = insert(root, 20)
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root.left.right
succ = inOrderSuccessor( root, temp)
if succ is not None:
print("Inorder Successor of" ,
temp.data ,"is" ,succ.data)
else:
print("InInorder Successor doesn't exist")
C#




// C# program for above approach
using System;
public class GFG
{
/* A binary tree node has data,
the pointer to left child
and a pointer to right child */
public
class node
{
public
int data;
public
node left;
public
node right;
public
node parent;
};
static node inOrderSuccessor(
node root,
node n)
{
// step 1 of the above algorithm
if (n.right != null)
return minValue(n.right);
node succ = null;
// Start from root and search for
// successor down the tree
while (root != null)
{
if (n.data < root.data)
{
succ = root;
root = root.left;
}
else if (n.data > root.data)
root = root.right;
else
break;
}
return succ;
}
/* Given a non-empty binary search tree,
return the minimum data
value found in that tree. Note that
the entire tree does not need
to be searched. */
static node minValue(node node)
{
node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null)
{
current = current.left;
}
return current;
}
/* Helper function that allocates a new
node with the given data and
null left and right pointers. */
static node newNode(int data)
{
node node = new node();
node.data = data;
node.left = null;
node.right = null;
node.parent = null;
return (node);
}
/* Give a binary search tree and
a number, inserts a new node with
the given number in the correct
place in the tree. Returns the new
root pointer which the caller should
then use (the standard trick to
astatic void using reference parameters). */
static node insert(node node,
int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null)
return (newNode(data));
else
{
node temp;
/* 2. Otherwise, recur down the tree */
if (data <= node.data)
{
temp = insert(node.left, data);
node.left = temp;
temp.parent = node;
}
else
{
temp = insert(node.right, data);
node.right = temp;
temp.parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
/* Driver program to test above functions*/
public static void Main(String[] args)
{
node root = null, temp, succ;
// creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root.left.right.right;
// Function Call
succ = inOrderSuccessor(root, temp);
if (succ != null)
Console.Write(
"\n Inorder Successor of {0} is {1} ",
temp.data, succ.data);
else
Console.Write("\n Inorder Successor doesn't exit");
}
}
// This code is contributed by gauravrajput1
Javascript




<script>
class Node
{
constructor(data)
{
this.data=data;;
this.left=this.right=this.parent=null;
}
}
function inOrderSuccessor(root,n)
{
// step 1 of the above algorithm
if (n.right != null)
return minValue(n.right);
let succ = null;
// Start from root and search for
// successor down the tree
while (root != null)
{
if (n.data < root.data)
{
succ = root;
root = root.left;
}
else if (n.data > root.data)
root = root.right;
else
break;
}
return succ;
}
function minValue(node)
{
let current = node;
/* loop down to find the leftmost leaf */
while (current.left != null)
{
current = current.left;
}
return current;
}
function insert(node,data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null)
return (new Node(data));
else
{
let temp;
/* 2. Otherwise, recur down the tree */
if (data <= node.data)
{
temp = insert(node.left, data);
node.left = temp;
temp.parent = node;
}
else
{
temp = insert(node.right, data);
node.right = temp;
temp.parent = node;
}
/* return the (unchanged) node pointer */
return node;
}
}
let root = null, temp, succ, min;
// creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root.left.right.right;
// Function Call
succ = inOrderSuccessor(root, temp);
if (succ != null)
document.write(
"<br> Inorder Successor of "+temp.data+" is "+
succ.data);
else
document.write("<br> Inorder Successor doesn't exit");
// This code is contributed by unknown2108
</script>
Output Inorder Successor of 14 is 20

Complexity Analysis:

  • Time Complexity: O(h), where h is the height of the tree.
    In the worst case as explained above we travel the whole height of the tree
  • Auxiliary Space: O(1).
    Due to no use of any data structure for storing values.

Method 3 (Inorder traversal)An inorder transversal of BST produces a sorted sequence. Therefore, we perform an inorder traversal. The first encountered node with value greater than the node is the inorder successor.

Input: node, root // node is the node whose ignorer successor is needed.
Output: succ // succ is Inorder successor of node.

Below is the implementation of the above approach:

C++




// C++ program for above approach
#include <iostream>
using namespace std;
/* A binary tree node has data,
the pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
struct node* parent;
};
struct node* newNode(int data);
void inOrderTraversal(struct node* root,
struct node* n,
struct node* succ)
{
if(root==nullptr) { return; }
inOrderTraversal(root->left, n, succ);
if(root->data>n->data && !succ->left) { succ->left = root; return; }
inOrderTraversal(root->right, n, succ);
}
struct node* inOrderSuccessor(struct node* root,
struct node* n)
{
struct node* succ = newNode(0);
inOrderTraversal(root, n, succ);
return succ->left;
}
// Helper function that allocates a new
// node with the given data and NULL left
// and right pointers.
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
node->parent = NULL;
return (node);
}
// Give a binary search tree and a
// number, inserts a new node with
// the given number in the correct
// place in the tree. Returns the new
// root pointer which the caller should
// then use (the standard trick to
// avoid using reference parameters).
struct node* insert(struct node* node,
int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return (newNode(data));
else
{
struct node* temp;
/* 2. Otherwise, recur down the tree */
if (data <= node->data)
{
temp = insert(node->left, data);
node->left = temp;
temp->parent = node;
}
else
{
temp = insert(node->right, data);
node->right = temp;
temp->parent = node;
}
/* Return the (unchanged) node pointer */
return node;
}
}
// Driver code
int main()
{
struct node *root = NULL, *temp, *succ, *min;
// Creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root->left->right->right;
// Function Call
succ = inOrderSuccessor(root, temp);
if (succ != NULL)
cout << "\n Inorder Successor of "
<< temp->data << " is "<< succ->data;
else
cout <<"\n Inorder Successor doesn't exist";
//getchar();
return 0;
}
// This code is contributed by jaisw7
Java




// Java program for above approach
import java.util.*;
class GFG {
/*
* A binary tree node has data, the pointer to left child and a pointer to right
* child
*/
static class node {
int data;
node left;
node right;
node parent;
};
static void inOrderTraversal(node root) {
if (root == null) {
return;
}
inOrderTraversal(root.left);
System.out.print(root.data);
inOrderTraversal(root.right);
}
static void inOrderTraversal(node root, node n, node succ) {
if (root == null) {
return;
}
inOrderTraversal(root.left, n, succ);
if (root.data > n.data && succ.left == null) {
succ.left = root;
return;
}
inOrderTraversal(root.right, n, succ);
}
static node inOrderSuccessor(node root, node n) {
node succ = newNode(0);
inOrderTraversal(root, n, succ);
return succ.left;
}
// Helper function that allocates a new
// node with the given data and null left
// and right pointers.
static node newNode(int data) {
node node = new node();
node.data = data;
node.left = null;
node.right = null;
node.parent = null;
return (node);
}
// Give a binary search tree and a
// number, inserts a new node with
// the given number in the correct
// place in the tree. Returns the new
// root pointer which the caller should
// then use (the standard trick to
// astatic void using reference parameters).
static node insert(node node, int data) {
/*
* 1. If the tree is empty, return a new, single node
*/
if (node == null)
return (newNode(data));
else {
node temp;
/* 2. Otherwise, recur down the tree */
if (data <= node.data) {
temp = insert(node.left, data);
node.left = temp;
temp.parent = node;
} else {
temp = insert(node.right, data);
node.right = temp;
temp.parent = node;
}
/* Return the (unchanged) node pointer */
return node;
}
}
// Driver code
public static void main(String[] args) {
node root = null, temp, succ, min;
// Creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root.left.right.right;
// Function Call
succ = inOrderSuccessor(root, temp);
if (succ != null)
System.out.print("\n Inorder Successor of " + temp.data + " is " + succ.data);
else
System.out.print("\n Inorder Successor doesn't exist");
}
}
// This code is contributed by Rajput-Ji
C#




// C# program for above approach
using System;
using System.Collections.Generic;
public class GFG {
/*
* A binary tree node has data, the pointer to left child and a pointer to right
* child
*/
public class node {
public int data;
public node left;
public node right;
public node parent;
};
static void inOrderTraversal(node root) {
if (root == null) {
return;
}
inOrderTraversal(root.left);
Console.Write(root.data);
inOrderTraversal(root.right);
}
static void inOrderTraversal(node root, node n, node succ) {
if (root == null) {
return;
}
inOrderTraversal(root.left, n, succ);
if (root.data > n.data && succ.left == null) {
succ.left = root;
return;
}
inOrderTraversal(root.right, n, succ);
}
static node inOrderSuccessor(node root, node n) {
node succ = newNode(0);
inOrderTraversal(root, n, succ);
return succ.left;
}
// Helper function that allocates a new
// node with the given data and null left
// and right pointers.
static node newNode(int data) {
node node = new node();
node.data = data;
node.left = null;
node.right = null;
node.parent = null;
return (node);
}
// Give a binary search tree and a
// number, inserts a new node with
// the given number in the correct
// place in the tree. Returns the new
// root pointer which the caller should
// then use (the standard trick to
// astatic void using reference parameters).
static node insert(node node, int data) {
/*
* 1. If the tree is empty, return a new, single node
*/
if (node == null)
return (newNode(data));
else {
node temp;
/* 2. Otherwise, recur down the tree */
if (data <= node.data) {
temp = insert(node.left, data);
node.left = temp;
temp.parent = node;
} else {
temp = insert(node.right, data);
node.right = temp;
temp.parent = node;
}
/* Return the (unchanged) node pointer */
return node;
}
}
// Driver code
public static void Main(String[] args) {
node root = null, temp, succ, min;
// Creating the tree given in the above diagram
root = insert(root, 20);
root = insert(root, 8);
root = insert(root, 22);
root = insert(root, 4);
root = insert(root, 12);
root = insert(root, 10);
root = insert(root, 14);
temp = root.left.right.right;
// Function Call
succ = inOrderSuccessor(root, temp);
if (succ != null)
Console.Write("\n Inorder Successor of " + temp.data + " is " + succ.data);
else
Console.Write("\n Inorder Successor doesn't exist");
}
}
// This code contributed by Rajput-Ji
Output Inorder Successor of 14 is 20

Complexity Analysis:

Time Complexity: O(h), where h is the height of the tree.In the worst case as explained above we travel the whole height of the tree
Auxiliary Space: O(1).Due to no use of any data structure for storing values.

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References:
//net.pku.edu.cn/~course/cs101/2007/resource/Intro2Algorithm/book6/chap13.htm
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.




Article Tags :
Binary Search Tree
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Morgan Stanley
Practice Tags :
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Binary Search Tree
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Linked List

  • Last Updated : 29 Sep, 2020

1234

Basic concepts and nomenclatureEdit

Each record of a linked list is often called an 'element' or 'node'.

The field of each node that contains the address of the next node is usually called the 'next link' or 'next pointer'. The remaining fields are known as the 'data', 'information', 'value', 'cargo', or 'payload' fields.

The 'head' of a list is its first node. The 'tail' of a list may refer either to the rest of the list after the head, or to the last node in the list. In Lisp and some derived languages, the next node may be called the 'cdr' (pronounced could-er) of the list, while the payload of the head node may be called the 'car'.

Singly linked listEdit

Singly linked lists contain nodes which have a data field as well as 'next' field, which points to the next node in line of nodes. Operations that can be performed on singly linked lists include insertion, deletion and traversal.

A singly linked list whose nodes contain two fields: an integer value and a link to the next node

The following code demonstrates how to add a new node with data "value" to the end of a singly linked list:

node addNode(node head, int value) { node temp, p; // declare two nodes temp and p temp = createNode(); // assume createNode creates a new node with data = 0 and next pointing to NULL. temp->data = value; // add element's value to data part of node if (head == NULL) { head = temp; // when linked list is empty } else { p = head; // assign head to p while (p->next != NULL) { p = p->next; // traverse the list until p is the last node. The last node always points to NULL. } p->next = temp; // Point the previous last node to the new node created. } return head; }

Doubly linked listEdit

In a 'doubly linked list', each node contains, besides the next-node link, a second link field pointing to the 'previous' node in the sequence. The two links may be called 'forward('s') and 'backwards', or 'next' and 'prev'('previous').

A doubly linked list whose nodes contain three fields: an integer value, the link forward to the next node, and the link backward to the previous node

A technique known as XOR-linking allows a doubly linked list to be implemented using a single link field in each node. However, this technique requires the ability to do bit operations on addresses, and therefore may not be available in some high-level languages.

Many modern operating systems use doubly linked lists to maintain references to active processes, threads, and other dynamic objects.[2] A common strategy for rootkits to evade detection is to unlink themselves from these lists.[3]

Multiply linked listEdit

In a 'multiply linked list', each node contains two or more link fields, each field being used to connect the same set of data records in a different order of same set (e.g., by name, by department, by date of birth, etc.). While doubly linked lists can be seen as special cases of multiply linked list, the fact that the two and more orders are opposite to each other leads to simpler and more efficient algorithms, so they are usually treated as a separate case.

Circular linked listEdit

In the last node of a list, the link field often contains a null reference, a special value is used to indicate the lack of further nodes. A less common convention is to make it point to the first node of the list; in that case, the list is said to be 'circular' or 'circularly linked'; otherwise, it is said to be 'open' or 'linear'. It is a list where the last pointer points to the first node.

In the case of a circular doubly linked list, the first node also points to the last node of the list.

Sentinel nodesEdit

In some implementations an extra 'sentinel' or 'dummy' node may be added before the first data record or after the last one. This convention simplifies and accelerates some list-handling algorithms, by ensuring that all links can be safely dereferenced and that every list (even one that contains no data elements) always has a "first" and "last" node.

Empty listsEdit

An empty list is a list that contains no data records. This is usually the same as saying that it has zero nodes. If sentinel nodes are being used, the list is usually said to be empty when it has only sentinel nodes.

Hash linkingEdit

The link fields need not be physically part of the nodes. If the data records are stored in an array and referenced by their indices, the link field may be stored in a separate array with the same indices as the data records.

List handlesEdit

Since a reference to the first node gives access to the whole list, that reference is often called the 'address', 'pointer', or 'handle' of the list. Algorithms that manipulate linked lists usually get such handles to the input lists and return the handles to the resulting lists. In fact, in the context of such algorithms, the word "list" often means "list handle". In some situations, however, it may be convenient to refer to a list by a handle that consists of two links, pointing to its first and last nodes.

Combining alternativesEdit

The alternatives listed above may be arbitrarily combined in almost every way, so one may have circular doubly linked lists without sentinels, circular singly linked lists with sentinels, etc.

TradeoffsEdit

As with most choices in computer programming and design, no method is well suited to all circumstances. A linked list data structure might work well in one case, but cause problems in another. This is a list of some of the common tradeoffs involving linked list structures.

Linked lists vs. dynamic arraysEdit

A dynamic array is a data structure that allocates all elements contiguously in memory, and keeps a count of the current number of elements. If the space reserved for the dynamic array is exceeded, it is reallocated and (possibly) copied, which is an expensive operation.

Linked lists have several advantages over dynamic arrays. Insertion or deletion of an element at a specific point of a list, assuming that we have indexed a pointer to the node (before the one to be removed, or before the insertion point) already, is a constant-time operation (otherwise without this reference it is O(n)), whereas insertion in a dynamic array at random locations will require moving half of the elements on average, and all the elements in the worst case. While one can "delete" an element from an array in constant time by somehow marking its slot as "vacant", this causes fragmentation that impedes the performance of iteration.

Moreover, arbitrarily many elements may be inserted into a linked list, limited only by the total memory available; while a dynamic array will eventually fill up its underlying array data structure and will have to reallocate—an expensive operation, one that may not even be possible if memory is fragmented, although the cost of reallocation can be averaged over insertions, and the cost of an insertion due to reallocation would still be amortized O(1). This helps with appending elements at the array's end, but inserting into (or removing from) middle positions still carries prohibitive costs due to data moving to maintain contiguity. An array from which many elements are removed may also have to be resized in order to avoid wasting too much space.

On the other hand, dynamic arrays (as well as fixed-size array data structures) allow constant-time random access, while linked lists allow only sequential access to elements. Singly linked lists, in fact, can be easily traversed in only one direction. This makes linked lists unsuitable for applications where it's useful to look up an element by its index quickly, such as heapsort. Sequential access on arrays and dynamic arrays is also faster than on linked lists on many machines, because they have optimal locality of reference and thus make good use of data caching.

Another disadvantage of linked lists is the extra storage needed for references, which often makes them impractical for lists of small data items such as characters or boolean values, because the storage overhead for the links may exceed by a factor of two or more the size of the data. In contrast, a dynamic array requires only the space for the data itself (and a very small amount of control data).[note 1] It can also be slow, and with a naïve allocator, wasteful, to allocate memory separately for each new element, a problem generally solved using memory pools.

Some hybrid solutions try to combine the advantages of the two representations. Unrolled linked lists store several elements in each list node, increasing cache performance while decreasing memory overhead for references. CDR coding does both these as well, by replacing references with the actual data referenced, which extends off the end of the referencing record.

A good example that highlights the pros and cons of using dynamic arrays vs. linked lists is by implementing a program that resolves the Josephus problem. The Josephus problem is an election method that works by having a group of people stand in a circle. Starting at a predetermined person, one may count around the circle n times. Once the nth person is reached, one should remove them from the circle and have the members close the circle. The process is repeated until only one person is left. That person wins the election. This shows the strengths and weaknesses of a linked list vs. a dynamic array, because if the people are viewed as connected nodes in a circular linked list, then it shows how easily the linked list is able to delete nodes (as it only has to rearrange the links to the different nodes). However, the linked list will be poor at finding the next person to remove and will need to search through the list until it finds that person. A dynamic array, on the other hand, will be poor at deleting nodes (or elements) as it cannot remove one node without individually shifting all the elements up the list by one. However, it is exceptionally easy to find the nth person in the circle by directly referencing them by their position in the array.

The list ranking problem concerns the efficient conversion of a linked list representation into an array. Although trivial for a conventional computer, solving this problem by a parallel algorithm is complicated and has been the subject of much research.

A balanced tree has similar memory access patterns and space overhead to a linked list while permitting much more efficient indexing, taking O(log n) time instead of O(n) for a random access. However, insertion and deletion operations are more expensive due to the overhead of tree manipulations to maintain balance. Schemes exist for trees to automatically maintain themselves in a balanced state: AVL trees or red–black trees.

Singly linked linear lists vs. other listsEdit

While doubly linked and circular lists have advantages over singly linked linear lists, linear lists offer some advantages that make them preferable in some situations.

A singly linked linear list is a recursive data structure, because it contains a pointer to a smaller object of the same type. For that reason, many operations on singly linked linear lists (such as merging two lists, or enumerating the elements in reverse order) often have very simple recursive algorithms, much simpler than any solution using iterative commands. While those recursive solutions can be adapted for doubly linked and circularly linked lists, the procedures generally need extra arguments and more complicated base cases.

Linear singly linked lists also allow tail-sharing, the use of a common final portion of sub-list as the terminal portion of two different lists. In particular, if a new node is added at the beginning of a list, the former list remains available as the tail of the new one—a simple example of a persistent data structure. Again, this is not true with the other variants: a node may never belong to two different circular or doubly linked lists.

In particular, end-sentinel nodes can be shared among singly linked non-circular lists. The same end-sentinel node may be used for every such list. In Lisp, for example, every proper list ends with a link to a special node, denoted by nil or (), whose CAR and CDR links point to itself. Thus a Lisp procedure can safely take the CAR or CDR of any list.

The advantages of the fancy variants are often limited to the complexity of the algorithms, not in their efficiency. A circular list, in particular, can usually be emulated by a linear list together with two variables that point to the first and last nodes, at no extra cost.

Doubly linked vs. singly linkedEdit

Double-linked lists require more space per node (unless one uses XOR-linking), and their elementary operations are more expensive; but they are often easier to manipulate because they allow fast and easy sequential access to the list in both directions. In a doubly linked list, one can insert or delete a node in a constant number of operations given only that node's address. To do the same in a singly linked list, one must have the address of the pointer to that node, which is either the handle for the whole list (in case of the first node) or the link field in the previous node. Some algorithms require access in both directions. On the other hand, doubly linked lists do not allow tail-sharing and cannot be used as persistent data structures.

Circularly linked vs. linearly linkedEdit

A circularly linked list may be a natural option to represent arrays that are naturally circular, e.g. the corners of a polygon, a pool of buffers that are used and released in FIFO ("first in, first out") order, or a set of processes that should be time-shared in round-robin order. In these applications, a pointer to any node serves as a handle to the whole list.

With a circular list, a pointer to the last node gives easy access also to the first node, by following one link. Thus, in applications that require access to both ends of the list (e.g., in the implementation of a queue), a circular structure allows one to handle the structure by a single pointer, instead of two.

A circular list can be split into two circular lists, in constant time, by giving the addresses of the last node of each piece. The operation consists in swapping the contents of the link fields of those two nodes. Applying the same operation to any two nodes in two distinct lists joins the two list into one. This property greatly simplifies some algorithms and data structures, such as the quad-edge and face-edge.

The simplest representation for an empty circular list (when such a thing makes sense) is a null pointer, indicating that the list has no nodes. Without this choice, many algorithms have to test for this special case, and handle it separately. By contrast, the use of null to denote an empty linear list is more natural and often creates fewer special cases.

For some applications, it can be useful to use singly linked lists that can vary between being circular and being linear, or even circular with a linear initial segment. Algorithms for searching or otherwise operating on these have to take precautions to avoid accidentally entering an endless loop. One usual method is to have a second pointer walking the list at half or double the speed, and if both pointers meet at the same node, you know you found a cycle.

Using sentinel nodesEdit

Sentinel node may simplify certain list operations, by ensuring that the next or previous nodes exist for every element, and that even empty lists have at least one node. One may also use a sentinel node at the end of the list, with an appropriate data field, to eliminate some end-of-list tests. For example, when scanning the list looking for a node with a given value x, setting the sentinel's data field to x makes it unnecessary to test for end-of-list inside the loop. Another example is the merging two sorted lists: if their sentinels have data fields set to +∞, the choice of the next output node does not need special handling for empty lists.

However, sentinel nodes use up extra space (especially in applications that use many short lists), and they may complicate other operations (such as the creation of a new empty list).

However, if the circular list is used merely to simulate a linear list, one may avoid some of this complexity by adding a single sentinel node to every list, between the last and the first data nodes. With this convention, an empty list consists of the sentinel node alone, pointing to itself via the next-node link. The list handle should then be a pointer to the last data node, before the sentinel, if the list is not empty; or to the sentinel itself, if the list is empty.

The same trick can be used to simplify the handling of a doubly linked linear list, by turning it into a circular doubly linked list with a single sentinel node. However, in this case, the handle should be a single pointer to the dummy node itself.[8]

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