Do doubly linked lists need a head?

Doubly Linked List

Hello Everyone,

Today I will be talking about a data structure called Doubly Linked List. Before I start talking about the Doubly linked list, I highly recommend reading my blog on Singly Linked List (Click Here). It will give you an overview of the Singly Linked List data. What is Doubly Linked List? A Doubly Linked List (points to two directions) very identical to Singly Linked List except it contains an extra pointer that keeps track of the previous node. Doubly Linked List has two pointers on every node that keeps track of the previous node and next node, unlike Singly Linked List which only keeps track of the next node. Just like the Singly Linked List, the first node in the Doubly Linked List is also called the head and the last node is also called the tail. In Doubly Linked List each node stores three things, data (integer or string), a reference to the next node and a previous node.

Do doubly linked lists need a head?
Doubly Linked List

So in the picture above we have a head node (A) pointing to the next node (B) and then also has a previous pointer but there’s nothing there that’s why it's NULL. If we go to the node ( B ), you can see it's pointing to the next node ( C ) as well as it’s pointing to the previous node ( A ).

Now let’s implement the Doubly Linked List (JavaScript).

We need two classes one called Node, Other DoublyLinkedList.

  • A Node class will have value, next and prev.
  • A DoublyLinkedList class will have a head, tail, and length.
class Node {
constructor(val) {
this.val = val;
this.next = null;
this.prev = null;
}
}
class DoublyLinkedList {
constructor() {
this.head = null;
this.tail = null;
this.length = 0;
}
}

At the moment Doubly Linked List doesn't do anything right now. Let’s write a method called the push to add a node to the end of the Doubly Linked List.

class DoublyLinkedList {
constructor() {
this.head = null;
this.tail = null;
this.length = 0;
}
push(val) {
let newNode = new Node(val);
if (this.head === null) {
this.head = newNode;
this.tail = newNode;
} else {
this.tail.next = newNode;
newNode.prev = this.tail
this.tail = newNode;
}
this.length++
return this;
}

}
Do doubly linked lists need a head?
Push Method Visual From https://visualgo.net/en/list

Let’s Implement method Pop to remove from the end

class DoublyLinkedList {
constructor() {
this.head = null;
this.tail = null;
this.length = 0;
}
push(val) {
let newNode = new Node(val);
if (this.head === null) {
this.head = newNode;
this.tail = newNode;
} else {
this.tail.next = newNode;
newNode.prev = this.tail
this.tail = newNode;
}
this.length++
return this;
}
pop() {
if (!this.head)
return undefined;
let currentTail = this.tail;
if (this.length === 1) {
this.head = null
this.tail = null
} else {
this.tail = currentTail.prev;
this.tail.next = null;
currentTail.prev = null;
}
this.length--;
return currentTail;
}

}
Do doubly linked lists need a head?
Pop Method Visual From https://visualgo.net/en/list

Next method Shift to remove from the beginning of the Doubly Linked List

class DoublyLinkedList {
constructor() {
this.head = null;
this.tail = null;
this.length = 0;
}
push(val) {
let newNode = new Node(val);
if (this.head === null) {
this.head = newNode;
this.tail = newNode;
} else {
this.tail.next = newNode;
newNode.prev = this.tail
this.tail = newNode;
}
this.length++
return this;
}
pop() {
if (!this.head)
return undefined;
let currentTail = this.tail;
if (this.length === 1) {
this.head = null
this.tail = null
} else {
this.tail = currentTail.prev;
this.tail.next = null;
currentTail.prev = null;
}
this.length--;
return currentTail;
}
shift() {
let oldHead = this.head;
if (this.length === 0)
return undefined;
else if (this.length === 1) {
this.head = null;
this.tail = null;
} else {
this.head = oldHead.next;
this.head.prev = null;
oldHead.next = null;
}
this.length--;
return oldHead;
}

}
Do doubly linked lists need a head?
Shift Method Visual From https://visualgo.net/en/list

Finally, a method that adds in the front called unshift.

class DoublyLinkedList {
constructor() {
this.head = null;
this.tail = null;
this.length = 0;
}
push(val) {
let newNode = new Node(val);
if (this.head === null) {
this.head = newNode;
this.tail = newNode;
} else {
this.tail.next = newNode;
newNode.prev = this.tail
this.tail = newNode;
}
this.length++
return this;
}
pop() {
if (!this.head)
return undefined;
let currentTail = this.tail;
if (this.length === 1) {
this.head = null
this.tail = null
} else {
this.tail = currentTail.prev;
this.tail.next = null;
currentTail.prev = null;
}
this.length--;
return currentTail;
}
shift() {
let oldHead = this.head;
if (this.length === 0)
return undefined;
else if (this.length === 1) {
this.head = null;
this.tail = null;
} else {
this.head = oldHead.next;
this.head.prev = null;
oldHead.next = null;
}
this.length--;
return oldHead;
}
unshift(val) {
let newNode = new Node(val);
if (this.length === 0) {
this.head = newNode;
this.tail = newNode;
} else {
this.head.prev = newNode;
newNode.next = this.head;
this.head = newNode;
}
this.length++
return this
}

}
Do doubly linked lists need a head?
UnShift Method Visual From https://visualgo.net/en/list
  • Check out my GitHub repository for the code above HERE, it also contains many more methods.
  • Gif Animations HERE

There you have it. I hope you found this blog helpful. If you're having any difficulty or need more explanation, I’m more than happy to help you. Please connect with me on LinkedIn or my email is .

Happy Coding :)

Do doubly linked lists need a head?
Src: https://s3.amazonaws.com/msoe/files/callouts/wide_sml_computer-science-landing-page.jpg

References:

Doubly Linked List | Set 1 (Introduction and Insertion)

We strongly recommend to refer following post as a prerequisite of this post.
Linked List Introduction
Inserting a node in Singly Linked List
A Doubly Linked List (DLL) contains an extra pointer, typically called previous pointer, together with next pointer and data which are there in singly linked list.

Do doubly linked lists need a head?

Following is representation of a DLL node in C language.




/* Node of a doubly linked list */
class Node
{
public:
int data;
Node* next; // Pointer to next node in DLL
Node* prev; // Pointer to previous node in DLL
};
// This code is contributed by shivanisinghss2110




/* Node of a doubly linked list */
struct Node {
int data;
struct Node* next; // Pointer to next node in DLL
struct Node* prev; // Pointer to previous node in DLL
};




// Class for Doubly Linked List
public class DLL {
Node head; // head of list
/* Doubly Linked list Node*/
class Node {
int data;
Node prev;
Node next;
// Constructor to create a new node
// next and prev is by default initialized as null
Node(int d) { data = d; }
}
}




# Node of a doubly linked list
class Node:
def __init__(self, next=None, prev=None, data=None):
self.next = next # reference to next node in DLL
self.prev = prev # reference to previous node in DLL
self.data = data




// Class for Doubly Linked List
public class DLL {
Node head; // head of list
/* Doubly Linked list Node*/
public class Node {
public int data;
public Node prev;
public Node next;
// Constructor to create a new node
// next and prev is by default initialized as null
Node(int d) { data = d; }
}
}
// This code contributed by gauravrajput1




<script>
// Class for Doubly Linked List
var head; // head of list
/* Doubly Linked list Node */
class Node {
// Constructor to create a new node
// next and prev is by default initialized as null
constructor(val) {
this.data = val;
this.prev = null;
this.next = null;
}
}
// This code contributed by gauravrajput1
</script>

Following are advantages/disadvantages of doubly linked list over singly linked list.
Advantages over singly linked list
1) A DLL can be traversed in both forward and backward direction.
2) The delete operation in DLL is more efficient if pointer to the node to be deleted is given.
3) We can quickly insert a new node before a given node.
In singly linked list, to delete a node, pointer to the previous node is needed. To get this previous node, sometimes the list is traversed. In DLL, we can get the previous node using previous pointer.

Disadvantages over singly linked list
1) Every node of DLL Require extra space for an previous pointer. It is possible to implement DLL with single pointer though (See this and this).
2) All operations require an extra pointer previous to be maintained. For example, in insertion, we need to modify previous pointers together with next pointers. For example in following functions for insertions at different positions, we need 1 or 2 extra steps to set previous pointer.
Insertion
A node can be added in four ways
1) At the front of the DLL
2) After a given node.
3) At the end of the DLL
4) Before a given node.



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

1) Add a node at the front: (A 5 steps process)
The new node is always added before the head of the given Linked List. And newly added node becomes the new head of DLL. For example if the given Linked List is 10152025 and we add an item 5 at the front, then the Linked List becomes 510152025. Let us call the function that adds at the front of the list is push(). The push() must receive a pointer to the head pointer, because push must change the head pointer to point to the new node (See this)

Do doubly linked lists need a head?

Following are the 5 steps to add node at the front.




/* Given a reference (pointer to pointer)
to the head of a list
and an int, inserts a new node on the
front of the list. */
void push(Node** head_ref, int new_data)
{
/* 1. allocate node */
Node* new_node = new Node();
/* 2. put in the data */
new_node->data = new_data;
/* 3. Make next of new node as head
and previous as NULL */
new_node->next = (*head_ref);
new_node->prev = NULL;
/* 4. change prev of head node to new node */
if ((*head_ref) != NULL)
(*head_ref)->prev = new_node;
/* 5. move the head to point to the new node */
(*head_ref) = new_node;
}
// This code is contributed by shivanisinghss2110




/* Given a reference (pointer to pointer) to the head of a list
and an int, inserts a new node on the front of the list. */
void push(struct Node** head_ref, int new_data)
{
/* 1. allocate node */
struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
/* 2. put in the data */
new_node->data = new_data;
/* 3. Make next of new node as head and previous as NULL */
new_node->next = (*head_ref);
new_node->prev = NULL;
/* 4. change prev of head node to new node */
if ((*head_ref) != NULL)
(*head_ref)->prev = new_node;
/* 5. move the head to point to the new node */
(*head_ref) = new_node;
}




// Adding a node at the front of the list
public void push(int new_data)
{
/* 1. allocate node
* 2. put in the data */
Node new_Node = new Node(new_data);
/* 3. Make next of new node as head and previous as NULL */
new_Node.next = head;
new_Node.prev = null;
/* 4. change prev of head node to new node */
if (head != null)
head.prev = new_Node;
/* 5. move the head to point to the new node */
head = new_Node;
}




# Adding a node at the front of the list
def push(self, new_data):
# 1 & 2: Allocate the Node & Put in the data
new_node = Node(data = new_data)
# 3. Make next of new node as head and previous as NULL
new_node.next = self.head
new_node.prev = None
# 4. change prev of head node to new node
if self.head is not None:
self.head.prev = new_node
# 5. move the head to point to the new node
self.head = new_node
# This code is contributed by jatinreaper




// Adding a node at the front of the list
public void push(int new_data)
{
/* 1. allocate node
* 2. put in the data */
Node new_Node = new Node(new_data);
/* 3. Make next of new node as head and previous as NULL */
new_Node.next = head;
new_Node.prev = null;
/* 4. change prev of head node to new node */
if (head != null)
head.prev = new_Node;
/* 5. move the head to point to the new node */
head = new_Node;
}
// This code is contributed by aashish2995




// Adding a node at the front of the list
function push(new_data)
{
/* 1. allocate node
* 2. put in the data */
let new_Node = new Node(new_data);
/* 3. Make next of new node as head and previous as NULL */
new_Node.next = head;
new_Node.prev = null;
/* 4. change prev of head node to new node */
if (head != null)
head.prev = new_Node;
/* 5. move the head to point to the new node */
head = new_Node;
}
// This code is contributed by saurabh_jaiswal.

Four steps of the above five steps are same as the 4 steps used for inserting at the front in singly linked list. The only extra step is to change previous of head.
2) Add a node after a given node.: (A 7 steps process)
We are given pointer to a node as prev_node, and the new node is inserted after the given node.

Do doubly linked lists need a head?




/* Given a node as prev_node, insert
a new node after the given node */
void insertAfter(Node* prev_node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_node == NULL)
{
cout<<"the given previous node cannot be NULL";
return;
}
/* 2. allocate new node */
Node* new_node = new Node();
/* 3. put in the data */
new_node->data = new_data;
/* 4. Make next of new node as next of prev_node */
new_node->next = prev_node->next;
/* 5. Make the next of prev_node as new_node */
prev_node->next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node->prev = prev_node;
/* 7. Change previous of new_node's next node */
if (new_node->next != NULL)
new_node->next->prev = new_node;
}
// This code is contributed by shivanisinghss2110.




/* Given a node as prev_node, insert a new node after the given node */
void insertAfter(struct Node* prev_node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_node == NULL) {
printf("the given previous node cannot be NULL");
return;
}
/* 2. allocate new node */
struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
/* 3. put in the data */
new_node->data = new_data;
/* 4. Make next of new node as next of prev_node */
new_node->next = prev_node->next;
/* 5. Make the next of prev_node as new_node */
prev_node->next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node->prev = prev_node;
/* 7. Change previous of new_node's next node */
if (new_node->next != NULL)
new_node->next->prev = new_node;
}




/* Given a node as prev_node, insert a new node after the given node */
public void InsertAfter(Node prev_Node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_Node == null) {
System.out.println("The given previous node cannot be NULL ");
return;
}
/* 2. allocate node
* 3. put in the data */
Node new_node = new Node(new_data);
/* 4. Make next of new node as next of prev_node */
new_node.next = prev_Node.next;
/* 5. Make the next of prev_node as new_node */
prev_Node.next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node.prev = prev_Node;
/* 7. Change previous of new_node's next node */
if (new_node.next != null)
new_node.next.prev = new_node;
}




# Given a node as prev_node, insert
# a new node after the given node
def insertAfter(self, prev_node, new_data):
# 1. check if the given prev_node is NULL
if prev_node is None:
print("This node doesn't exist in DLL")
return
#2. allocate node & 3. put in the data
new_node = Node(data = new_data)
# 4. Make next of new node as next of prev_node
new_node.next = prev_node.next
# 5. Make the next of prev_node as new_node
prev_node.next = new_node
# 6. Make prev_node as previous of new_node
new_node.prev = prev_node
# 7. Change previous of new_node's next node */
if new_node.next is not None:
new_node.next.prev = new_node
# This code is contributed by jatinreaper




/* Given a node as prev_node, insert a new node after the given node */
public void InsertAfter(Node prev_Node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_Node == null) {
Console.WriteLine("The given previous node cannot be NULL ");
return;
}
/* 2. allocate node
* 3. put in the data */
Node new_node = new Node(new_data);
/* 4. Make next of new node as next of prev_node */
new_node.next = prev_Node.next;
/* 5. Make the next of prev_node as new_node */
prev_Node.next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node.prev = prev_Node;
/* 7. Change previous of new_node's next node */
if (new_node.next != null)
new_node.next.prev = new_node;
}
// This code is contributed by aashish2995




<script>
function InsertAfter(prev_Node,new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_Node == null) {
document.write("The given previous node cannot be NULL <br>");
return;
}
/* 2. allocate node
* 3. put in the data */
let new_node = new Node(new_data);
/* 4. Make next of new node as next of prev_node */
new_node.next = prev_Node.next;
/* 5. Make the next of prev_node as new_node */
prev_Node.next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node.prev = prev_Node;
/* 7. Change previous of new_node's next node */
if (new_node.next != null)
new_node.next.prev = new_node;
}
// This code is contributed by unknown2108
</script>

Five of the above steps step process are same as the 5 steps used for inserting after a given node in singly linked list. The two extra steps are needed to change previous pointer of new node and previous pointer of new node’s next node.
3) Add a node at the end: (7 steps process)
The new node is always added after the last node of the given Linked List. For example if the given DLL is 510152025 and we add an item 30 at the end, then the DLL becomes 51015202530.
Since a Linked List is typically represented by the head of it, we have to traverse the list till end and then change the next of last node to new node.

Do doubly linked lists need a head?

Following are the 7 steps to add node at the end.




/* Given a reference (pointer to pointer) to the head
of a DLL and an int, appends a new node at the end */
void append(Node** head_ref, int new_data)
{
/* 1. allocate node */
Node* new_node = new Node();
Node* last = *head_ref; /* used in step 5*/
/* 2. put in the data */
new_node->data = new_data;
/* 3. This new node is going to be the last node, so
make next of it as NULL*/
new_node->next = NULL;
/* 4. If the Linked List is empty, then make the new
node as head */
if (*head_ref == NULL)
{
new_node->prev = NULL;
*head_ref = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last->next != NULL)
last = last->next;
/* 6. Change the next of last node */
last->next = new_node;
/* 7. Make last node as previous of new node */
new_node->prev = last;
return;
}
// This code is contributed by shivanisinghss2110




/* Given a reference (pointer to pointer) to the head
of a DLL and an int, appends a new node at the end */
void append(struct Node** head_ref, int new_data)
{
/* 1. allocate node */
struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
struct Node* last = *head_ref; /* used in step 5*/
/* 2. put in the data */
new_node->data = new_data;
/* 3. This new node is going to be the last node, so
make next of it as NULL*/
new_node->next = NULL;
/* 4. If the Linked List is empty, then make the new
node as head */
if (*head_ref == NULL) {
new_node->prev = NULL;
*head_ref = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last->next != NULL)
last = last->next;
/* 6. Change the next of last node */
last->next = new_node;
/* 7. Make last node as previous of new node */
new_node->prev = last;
return;
}




// Add a node at the end of the list
void append(int new_data)
{
/* 1. allocate node
* 2. put in the data */
Node new_node = new Node(new_data);
Node last = head; /* used in step 5*/
/* 3. This new node is going to be the last node, so
* make next of it as NULL*/
new_node.next = null;
/* 4. If the Linked List is empty, then make the new
* node as head */
if (head == null) {
new_node.prev = null;
head = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last.next != null)
last = last.next;
/* 6. Change the next of last node */
last.next = new_node;
/* 7. Make last node as previous of new node */
new_node.prev = last;
}




# Add a node at the end of the DLL
def append(self, new_data):
# 1. allocate node 2. put in the data
new_node = Node(data = new_data)
last = self.head
# 3. This new node is going to be the
# last node, so make next of it as NULL
new_node.next = None
# 4. If the Linked List is empty, then
# make the new node as head
if self.head is None:
new_node.prev = None
self.head = new_node
return
# 5. Else traverse till the last node
while (last.next is not None):
last = last.next
# 6. Change the next of last node
last.next = new_node
# 7. Make last node as previous of new node */
new_node.prev = last
# This code is contributed by jatinreaper




// Add a node at the end of the list
void append(int new_data)
{
/* 1. allocate node
* 2. put in the data */
Node new_node = new Node(new_data);
Node last = head; /* used in step 5*/
/* 3. This new node is going
to be the last node, so
* make next of it as NULL*/
new_node.next = null;
/* 4. If the Linked List is empty,
then make the new * node as head */
if (head == null)
{
new_node.prev = null;
head = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last.next != null)
last = last.next;
/* 6. Change the next of last node */
last.next = new_node;
/* 7. Make last node as previous of new node */
new_node.prev = last;
}
// This code is contributed by shivanisinghss2110




<script>
// Add a node at the end of the list
function append(new_data)
{
/* 1. allocate node
* 2. put in the data */
var new_node = new Node(new_data);
var last = head; /* used in step 5*/
/* 3. This new node is going to be the last node, so
* make next of it as NULL*/
new_node.next = null;
/* 4. If the Linked List is empty, then make the new
* node as head */
if (head == null) {
new_node.prev = null;
head = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last.next != null)
last = last.next;
/* 6. Change the next of last node */
last.next = new_node;
/* 7. Make last node as previous of new node */
new_node.prev = last;
}
// This code is contributed by Rajput-Ji
</script>

Six of the above 7 steps are same as the 6 steps used for inserting after a given node in singly linked list. The one extra step is needed to change previous pointer of new node.
4) Add a node before a given node:

Steps
Let the pointer to this given node be next_node and the data of the new node to be added as new_data.

  1. Check if the next_node is NULL or not. If it’s NULL, return from the function because any new node can not be added before a NULL
  2. Allocate memory for the new node, let it be called new_node
  3. Set new_node->data = new_data
  4. Set the previous pointer of this new_node as the previous node of the next_node, new_node->prev = next_node->prev
  5. Set the previous pointer of the next_node as the new_node, next_node->prev = new_node
  6. Set the next pointer of this new_node as the next_node, new_node->next = next_node;
  7. If the previous node of the new_node is not NULL, then set the next pointer of this previous node as new_node, new_node->prev->next = new_node
  8. Else, if the prev of new_node is NULL, it will be the new head node. So, make (*head_ref) = new_node.

Do doubly linked lists need a head?

Below is the implementation of the above approach:

Code block

Output:

Created DLL is:

Traversal in forward Direction

9 1 5 7 6

Traversal in reverse direction

6 7 5 1 9

A complete working program to test above functions.
Following is complete program to test above functions.




// A complete working C++ program to
// demonstrate all insertion methods
#include <bits/stdc++.h>
using namespace std;
// A linked list node
class Node
{
public:
int data;
Node* next;
Node* prev;
};
/* Given a reference (pointer to pointer)
to the head of a list
and an int, inserts a new node on the
front of the list. */
void push(Node** head_ref, int new_data)
{
/* 1. allocate node */
Node* new_node = new Node();
/* 2. put in the data */
new_node->data = new_data;
/* 3. Make next of new node as head
and previous as NULL */
new_node->next = (*head_ref);
new_node->prev = NULL;
/* 4. change prev of head node to new node */
if ((*head_ref) != NULL)
(*head_ref)->prev = new_node;
/* 5. move the head to point to the new node */
(*head_ref) = new_node;
}
/* Given a node as prev_node, insert
a new node after the given node */
void insertAfter(Node* prev_node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_node == NULL)
{
cout<<"the given previous node cannot be NULL";
return;
}
/* 2. allocate new node */
Node* new_node = new Node();
/* 3. put in the data */
new_node->data = new_data;
/* 4. Make next of new node as next of prev_node */
new_node->next = prev_node->next;
/* 5. Make the next of prev_node as new_node */
prev_node->next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node->prev = prev_node;
/* 7. Change previous of new_node's next node */
if (new_node->next != NULL)
new_node->next->prev = new_node;
}
/* Given a reference (pointer to pointer) to the head
of a DLL and an int, appends a new node at the end */
void append(Node** head_ref, int new_data)
{
/* 1. allocate node */
Node* new_node = new Node();
Node* last = *head_ref; /* used in step 5*/
/* 2. put in the data */
new_node->data = new_data;
/* 3. This new node is going to be the last node, so
make next of it as NULL*/
new_node->next = NULL;
/* 4. If the Linked List is empty, then make the new
node as head */
if (*head_ref == NULL)
{
new_node->prev = NULL;
*head_ref = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last->next != NULL)
last = last->next;
/* 6. Change the next of last node */
last->next = new_node;
/* 7. Make last node as previous of new node */
new_node->prev = last;
return;
}
// This function prints contents of
// linked list starting from the given node
void printList(Node* node)
{
Node* last;
cout<<"\nTraversal in forward direction \n";
while (node != NULL)
{
cout<<" "<<node->data<<" ";
last = node;
node = node->next;
}
cout<<"\nTraversal in reverse direction \n";
while (last != NULL)
{
cout<<" "<<last->data<<" ";
last = last->prev;
}
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
Node* head = NULL;
// Insert 6. So linked list becomes 6->NULL
append(&head, 6);
// Insert 7 at the beginning. So
// linked list becomes 7->6->NULL
push(&head, 7);
// Insert 1 at the beginning. So
// linked list becomes 1->7->6->NULL
push(&head, 1);
// Insert 4 at the end. So linked
// list becomes 1->7->6->4->NULL
append(&head, 4);
// Insert 8, after 7. So linked
// list becomes 1->7->8->6->4->NULL
insertAfter(head->next, 8);
cout << "Created DLL is: ";
printList(head);
return 0;
}
// This is code is contributed by rathbhupendra




// A complete working C program to
// demonstrate all insertion
// methods
#include <stdio.h>
#include <stdlib.h>
// A linked list node
struct Node {
int data;
struct Node* next;
struct Node* prev;
};
/* Given a reference (pointer to pointer) to the head of a
list and an int, inserts a new node on the front of the
list. */
void push(struct Node** head_ref, int new_data)
{
/* 1. allocate node */
struct Node* new_node
= (struct Node*)malloc(sizeof(struct Node));
/* 2. put in the data */
new_node->data = new_data;
/* 3. Make next of new node as head and previous as NULL
*/
new_node->next = (*head_ref);
new_node->prev = NULL;
/* 4. change prev of head node to new node */
if ((*head_ref) != NULL)
(*head_ref)->prev = new_node;
/* 5. move the head to point to the new node */
(*head_ref) = new_node;
}
/* Given a node as prev_node, insert a new node after the
* given node */
void insertAfter(struct Node* prev_node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_node == NULL) {
printf("the given previous node cannot be NULL");
return;
}
/* 2. allocate new node */
struct Node* new_node
= (struct Node*)malloc(sizeof(struct Node));
/* 3. put in the data */
new_node->data = new_data;
/* 4. Make next of new node as next of prev_node */
new_node->next = prev_node->next;
/* 5. Make the next of prev_node as new_node */
prev_node->next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node->prev = prev_node;
/* 7. Change previous of new_node's next node */
if (new_node->next != NULL)
new_node->next->prev = new_node;
}
/* Given a reference (pointer to pointer) to the head
of a DLL and an int, appends a new node at the end */
void append(struct Node** head_ref, int new_data)
{
/* 1. allocate node */
struct Node* new_node
= (struct Node*)malloc(sizeof(struct Node));
struct Node* last = *head_ref; /* used in step 5*/
/* 2. put in the data */
new_node->data = new_data;
/* 3. This new node is going to be the last node, so
make next of it as NULL*/
new_node->next = NULL;
/* 4. If the Linked List is empty, then make the new
node as head */
if (*head_ref == NULL) {
new_node->prev = NULL;
*head_ref = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last->next != NULL)
last = last->next;
/* 6. Change the next of last node */
last->next = new_node;
/* 7. Make last node as previous of new node */
new_node->prev = last;
return;
}
// This function prints contents of linked list starting
// from the given node
void printList(struct Node* node)
{
struct Node* last;
printf("\nTraversal in forward direction \n");
while (node != NULL) {
printf(" %d ", node->data);
last = node;
node = node->next;
}
printf("\nTraversal in reverse direction \n");
while (last != NULL) {
printf(" %d ", last->data);
last = last->prev;
}
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
// Insert 6. So linked list becomes 6->NULL
append(&head, 6);
// Insert 7 at the beginning. So linked list becomes
// 7->6->NULL
push(&head, 7);
// Insert 1 at the beginning. So linked list becomes
// 1->7->6->NULL
push(&head, 1);
// Insert 4 at the end. So linked list becomes
// 1->7->6->4->NULL
append(&head, 4);
// Insert 8, after 7. So linked list becomes
// 1->7->8->6->4->NULL
insertAfter(head->next, 8);
printf("Created DLL is: ");
printList(head);
getchar();
return 0;
}




// A complete working Java program to demonstrate all
// Class for Doubly Linked List
public class DLL {
Node head; // head of list
/* Doubly Linked list Node*/
class Node {
int data;
Node prev;
Node next;
// Constructor to create a new node
// next and prev is by default initialized as null
Node(int d) { data = d; }
}
// Adding a node at the front of the list
public void push(int new_data)
{
/* 1. allocate node
* 2. put in the data */
Node new_Node = new Node(new_data);
/* 3. Make next of new node as head and previous as NULL */
new_Node.next = head;
new_Node.prev = null;
/* 4. change prev of head node to new node */
if (head != null)
head.prev = new_Node;
/* 5. move the head to point to the new node */
head = new_Node;
}
// Add a node before the given node
public void InsertBefore(Node next_node, int new_data)
{
/*Check if the given nx_node is NULL*/
if(next_node == null)
{
System.out.println("The given next node can not be NULL");
return;
}
//Allocate node, put in the data
Node new_node = new Node(new_data);
//Making prev of new node as prev of next node
new_node.prev = next_node.prev;
//Making prev of next node as new node
next_node.prev = new_node;
//Making next of new node as next node
new_node.next = next_node;
//Check if new node is added as head
if(new_node.prev != null)
new_node.prev.next = new_node;
else
head = new_node;
}
/* Given a node as prev_node, insert
a new node after the given node */
public void InsertAfter(Node prev_Node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_Node == null) {
System.out.println("The given previous node cannot be NULL ");
return;
}
/* 2. allocate node
* 3. put in the data */
Node new_node = new Node(new_data);
/* 4. Make next of new node as next of prev_node */
new_node.next = prev_Node.next;
/* 5. Make the next of prev_node as new_node */
prev_Node.next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node.prev = prev_Node;
/* 7. Change previous of new_node's next node */
if (new_node.next != null)
new_node.next.prev = new_node;
}
// Add a node at the end of the list
void append(int new_data)
{
/* 1. allocate node
* 2. put in the data */
Node new_node = new Node(new_data);
Node last = head; /* used in step 5*/
/* 3. This new node is going to be the last node, so
* make next of it as NULL*/
new_node.next = null;
/* 4. If the Linked List is empty, then make the new
* node as head */
if (head == null) {
new_node.prev = null;
head = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last.next != null)
last = last.next;
/* 6. Change the next of last node */
last.next = new_node;
/* 7. Make last node as previous of new node */
new_node.prev = last;
}
// This function prints contents of
// linked list starting from the given node
public void printlist(Node node)
{
Node last = null;
System.out.println("Traversal in forward Direction");
while (node != null) {
System.out.print(node.data + " ");
last = node;
node = node.next;
}
System.out.println();
System.out.println("Traversal in reverse direction");
while (last != null) {
System.out.print(last.data + " ");
last = last.prev;
}
}
/* Driver program to test above functions*/
public static void main(String[] args)
{
/* Start with the empty list */
DLL dll = new DLL();
// Insert 6. So linked list becomes 6->NULL
dll.append(6);
// Insert 7 at the beginning. So
// linked list becomes 7->6->NULL
dll.push(7);
// Insert 1 at the beginning. So
// linked list becomes 1->7->6->NULL
dll.push(1);
// Insert 4 at the end. So linked
// list becomes 1->7->6->4->NULL
dll.append(4);
// Insert 8, after 7. So linked
// list becomes 1->7->8->6->4->NULL
dll.InsertAfter(dll.head.next, 8);
// Insert 5, before 8.So linked
// list becomes 1->7->5->8->6->4
dll.InsertBefore(dll.head.next.next, 5);
System.out.println("Created DLL is: ");
dll.printlist(dll.head);
}
}
// This code is contributed by Sumit Ghosh




# A complete working Python
# program to demonstrate all
# insertion methods
# A linked list node
class Node:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.next = None
self.prev = None
# Class to create a Doubly Linked List
class DoublyLinkedList:
# Constructor for empty Doubly Linked List
def __init__(self):
self.head = None
# Given a reference to the head of a list and an
# integer, inserts a new node on the front of list
def push(self, new_data):
# 1. Allocates node
# 2. Put the data in it
new_node = Node(new_data)
# 3. Make next of new node as head and
# previous as None (already None)
new_node.next = self.head
# 4. change prev of head node to new_node
if self.head is not None:
self.head.prev = new_node
# 5. move the head to point to the new node
self.head = new_node
# Given a node as prev_node, insert a new node after
# the given node
def insertAfter(self, prev_node, new_data):
# 1. Check if the given prev_node is None
if prev_node is None:
print("the given previous node cannot be NULL")
return
# 2. allocate new node
# 3. put in the data
new_node = Node(new_data)
# 4. Make net of new node as next of prev node
new_node.next = prev_node.next
# 5. Make prev_node as previous of new_node
prev_node.next = new_node
# 6. Make prev_node ass previous of new_node
new_node.prev = prev_node
# 7. Change previous of new_nodes's next node
if new_node.next:
new_node.next.prev = new_node
# Given a reference to the head of DLL and integer,
# appends a new node at the end
def append(self, new_data):
# 1. Allocates node
# 2. Put in the data
new_node = Node(new_data)
# 3. This new node is going to be the last node,
# so make next of it as None
# (It already is initialized as None)
# 4. If the Linked List is empty, then make the
# new node as head
if self.head is None:
self.head = new_node
return
# 5. Else traverse till the last node
last = self.head
while last.next:
last = last.next
# 6. Change the next of last node
last.next = new_node
# 7. Make last node as previous of new node
new_node.prev = last
return
# This function prints contents of linked list
# starting from the given node
def printList(self, node):
print("\nTraversal in forward direction")
while node:
print(" {}".format(node.data))
last = node
node = node.next
print("\nTraversal in reverse direction")
while last:
print(" {}".format(last.data))
last = last.prev
# Driver program to test above functions
# Start with empty list
llist = DoublyLinkedList()
# Insert 6. So the list becomes 6->None
llist.append(6)
# Insert 7 at the beginning.
# So linked list becomes 7->6->None
llist.push(7)
# Insert 1 at the beginning.
# So linked list becomes 1->7->6->None
llist.push(1)
# Insert 4 at the end.
# So linked list becomes 1->7->6->4->None
llist.append(4)
# Insert 8, after 7.
# So linked list becomes 1->7->8->6->4->None
llist.insertAfter(llist.head.next, 8)
print ("Created DLL is: ")
llist.printList(llist.head)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)




// A complete working C# program to demonstrate all
using System;
// Class for Doubly Linked List
public class DLL
{
Node head; // head of list
/* Doubly Linked list Node*/
public class Node
{
public int data;
public Node prev;
public Node next;
// Constructor to create a new node
// next and prev is by default initialized as null
public Node(int d)
{
data = d;
}
}
// Adding a node at the front of the list
public void push(int new_data)
{
/* 1. allocate node
* 2. put in the data */
Node new_Node = new Node(new_data);
/* 3. Make next of new node as
head and previous as NULL */
new_Node.next = head;
new_Node.prev = null;
/* 4. change prev of head node to new node */
if (head != null)
head.prev = new_Node;
/* 5. move the head to point to the new node */
head = new_Node;
}
/* Given a node as prev_node, insert
a new node after the given node */
public void InsertAfter(Node prev_Node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_Node == null)
{
Console.WriteLine("The given previous node cannot be NULL ");
return;
}
/* 2. allocate node
* 3. put in the data */
Node new_node = new Node(new_data);
/* 4. Make next of new node as next of prev_node */
new_node.next = prev_Node.next;
/* 5. Make the next of prev_node as new_node */
prev_Node.next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node.prev = prev_Node;
/* 7. Change previous of new_node's next node */
if (new_node.next != null)
new_node.next.prev = new_node;
}
// Add a node at the end of the list
void append(int new_data)
{
/* 1. allocate node
* 2. put in the data */
Node new_node = new Node(new_data);
Node last = head; /* used in step 5*/
/* 3. This new node is going
to be the last node, so
* make next of it as NULL*/
new_node.next = null;
/* 4. If the Linked List is empty,
then make the new * node as head */
if (head == null)
{
new_node.prev = null;
head = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last.next != null)
last = last.next;
/* 6. Change the next of last node */
last.next = new_node;
/* 7. Make last node as previous of new node */
new_node.prev = last;
}
// This function prints contents of
// linked list starting from the given node
public void printlist(Node node)
{
Node last = null;
Console.WriteLine("Traversal in forward Direction");
while (node != null) {
Console.Write(node.data + " ");
last = node;
node = node.next;
}
Console.WriteLine();
Console.WriteLine("Traversal in reverse direction");
while (last != null) {
Console.Write(last.data + " ");
last = last.prev;
}
}
/* Driver code*/
public static void Main(String[] args)
{
/* Start with the empty list */
DLL dll = new DLL();
// Insert 6. So linked list becomes 6->NULL
dll.append(6);
// Insert 7 at the beginning.
// So linked list becomes 7->6->NULL
dll.push(7);
// Insert 1 at the beginning.
// So linked list becomes 1->7->6->NULL
dll.push(1);
// Insert 4 at the end. So linked list
// becomes 1->7->6->4->NULL
dll.append(4);
// Insert 8, after 7. So linked list
// becomes 1->7->8->6->4->NULL
dll.InsertAfter(dll.head.next, 8);
Console.WriteLine("Created DLL is: ");
dll.printlist(dll.head);
}
}
// This code is contributed by 29AjayKumar




<script>
// A complete working javascript program to demonstrate all
// Class for Doubly Linked List
var head; // head of list
/* Doubly Linked list Node */
class Node {
// Constructor to create a new node
// next and prev is by default initialized as null
constructor(d) {
this.data = d;
this.next = null;
this.prev = null;
}
}
// Adding a node at the front of the list
function push(new_data) {
/*
* 1. allocate node 2. put in the data
*/
var new_Node = new Node(new_data);
/* 3. Make next of new node as head and previous as NULL */
new_Node.next = head;
new_Node.prev = null;
/* 4. change prev of head node to new node */
if (head != null)
head.prev = new_Node;
/* 5. move the head to point to the new node */
head = new_Node;
}
// Add a node before the given node
function InsertBefore(next_node , new_data) {
/* Check if the given nx_node is NULL */
if (next_node == null) {
document.write("The given next node can not be NULL");
return;
}
// Allocate node, put in the data
var new_node = new Node(new_data);
// Making prev of new node as prev of next node
new_node.prev = next_node.prev;
// Making prev of next node as new node
next_node.prev = new_node;
// Making next of new node as next node
new_node.next = next_node;
// Check if new node is added as head
if (new_node.prev != null)
new_node.prev.next = new_node;
else
head = new_node;
}
/*
* Given a node as prev_node, insert a new node after the given node
*/
function InsertAfter(prev_Node , new_data) {
/* 1. check if the given prev_node is NULL */
if (prev_Node == null) {
document.write("The given previous node cannot be NULL ");
return;
}
/*
* 2. allocate node 3. put in the data
*/
var new_node = new Node(new_data);
/* 4. Make next of new node as next of prev_node */
new_node.next = prev_Node.next;
/* 5. Make the next of prev_node as new_node */
prev_Node.next = new_node;
/* 6. Make prev_node as previous of new_node */
new_node.prev = prev_Node;
/* 7. Change previous of new_node's next node */
if (new_node.next != null)
new_node.next.prev = new_node;
}
// Add a node at the end of the list
function append(new_data) {
/*
* 1. allocate node 2. put in the data
*/
var new_node = new Node(new_data);
var last = head; /* used in step 5 */
/*
* 3. This new node is going to be the last node, so make next of it as NULL
*/
new_node.next = null;
/*
* 4. If the Linked List is empty, then make the new node as head
*/
if (head == null) {
new_node.prev = null;
head = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last.next != null)
last = last.next;
/* 6. Change the next of last node */
last.next = new_node;
/* 7. Make last node as previous of new node */
new_node.prev = last;
}
// This function prints contents of
// linked list starting from the given node
function printlist(node) {
var last = null;
document.write("<br/>Traversal in forward Direction<br/>");
while (node != null) {
document.write(node.data + " ");
last = node;
node = node.next;
}
document.write();
document.write("<br/>Traversal in reverse direction<br/>");
while (last != null) {
document.write(last.data + " ");
last = last.prev;
}
}
/* Driver program to test above functions */
/* Start with the empty list */
// Insert 6. So linked list becomes 6->NULL
append(6);
// Insert 7 at the beginning. So
// linked list becomes 7->6->NULL
push(7);
// Insert 1 at the beginning. So
// linked list becomes 1->7->6->NULL
push(1);
// Insert 4 at the end. So linked
// list becomes 1->7->6->4->NULL
append(4);
// Insert 8, after 7. So linked
// list becomes 1->7->8->6->4->NULL
InsertAfter(head.next, 8);
// Insert 5, before 8.So linked
// list becomes 1->7->5->8->6->4
InsertBefore(head.next.next, 5);
document.write("Created DLL is:<br/> ");
printlist(head);
// This code is contributed by Rajput-Ji
</script>

Output:

Created DLL is: Traversal in forward Direction 1 7 5 8 6 4 Traversal in reverse direction 4 6 8 5 7 1

Also see: Delete a node in double Link List
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Do doubly linked lists need a head?




Article Tags :
Data Structures
Linked List
doubly linked list
Visa
Practice Tags :
Visa
Data Structures
Linked List

5.6.1. Doubly Linked Lists

The singly linked list allows for direct access from a list node only to the next node in the list. A doubly linked list allows convenient access from a list node to the next node and also to the preceding node on the list. The doubly linked list node accomplishes this in the obvious way by storing two pointers: one to the node following it (as in the singly linked list), and a second pointer to the node preceding it.

Figure 5.6.1: A doubly linked list.

The most common reason to use a doubly linked list is because it is easier to implement than a singly linked list. While the code for the doubly linked implementation is a little longer than for the singly linked version, it tends to be a bit more “obvious” in its intention, and so easier to implement and debug. Whether a list implementation is doubly or singly linked should be hidden from the List class user.

Like our singly linked list implementation, the doubly linked list implementation makes use of a header node. We also add a tailer node to the end of the list. The tailer is similar to the header, in that it is a node that contains no value, and it always exists. When the doubly linked list is initialized, the header and tailer nodes are created. Data member head points to the header node, and tail points to the tailer node. The purpose of these nodes is to simplify the insert, append, and remove methods by eliminating all need for special-case code when the list is empty, or when we insert at the head or tail of the list.

In our implementation, curr will point to the current position (or to the trailer node if the current position is at the end of the list).

Here is the complete implementation for a Link class to be used with doubly linked lists. This code is a little longer than that for the singly linked list node implementation since the doubly linked list nodes have an extra data member.

class Link { // Doubly linked list node private Object e; // Value for this node private Link n; // Pointer to next node in list private Link p; // Pointer to previous node // Constructors Link(Object it, Link inp, Link inn) { e = it; p = inp; n = inn; } Link(Link inp, Link inn) { p = inp; n = inn; } // Get and set methods for the data members public Object element() { return e; } // Return the value public Object setElement(Object it) { return e = it; } // Set element value public Link next() { return n; } // Return next link public Link setNext(Link nextval) { return n = nextval; } // Set next link public Link prev() { return p; } // Return prev link public Link setPrev(Link prevval) { return p = prevval; } // Set prev link }
class Link { // Doubly linked list node private E e; // Value for this node private Link n; // Pointer to next node in list private Link p; // Pointer to previous node // Constructors Link(E it, Link inp, Link inn) { e = it; p = inp; n = inn; } Link(Link inp, Link inn) { p = inp; n = inn; } // Get and set methods for the data members public E element() { return e; } // Return the value public E setElement(E it) { return e = it; } // Set element value public Link next() { return n; } // Return next link public Link setNext(Link nextval) { return n = nextval; } // Set next link public Link prev() { return p; } // Return prev link public Link setPrev(Link prevval) { return p = prevval; } // Set prev link }

5.6.1.1. Insert

The following slideshows illustrate the insert and append doubly linked list methods. The class declaration and the remaining member functions for the doubly linked list class are nearly identical to the singly linked list version. While the code for these methods might be a little longer than their singly linked list counterparts (since there is an extra pointer in each node to deal with), they tend to be easier to understand.

5.6.1.4. Prev

The only disadvantage of the doubly linked list as compared to the singly linked list is the additional space used. The doubly linked list requires two pointers per node, and so in the implementation presented it requires twice as much overhead as the singly linked list.

5.6.1.5. Mangling Pointers

There is a space-saving technique that can be employed to eliminate the additional space requirement, though it will complicate the implementation and be somewhat slower. Thus, this is an example of a space/time tradeoff. It is based on observing that, if we store the sum of two values, then we can get either value back by subtracting the other. That is, if we store \(a + b\) in variable \(c\), then \(b = c - a\) and \(a = c - b\). Of course, to recover one of the values out of the stored summation, the other value must be supplied. A pointer to the first node in the list, along with the value of one of its two link fields, will allow access to all of the remaining nodes of the list in order. This is because the pointer to the node must be the same as the value of the following node’s prev pointer, as well as the previous node’s next pointer. It is possible to move down the list breaking apart the summed link fields as though you were opening a zipper.

The principle behind this technique is worth remembering, as it has many applications. The following code fragment will swap the contents of two variables without using a temporary variable (at the cost of three arithmetic operations).

a = a + b; b = a - b; // Now b contains original value of a a = a - b; // Now a contains original value of b
a = a + b; b = a - b; // Now b contains original value of a a = a - b; // Now a contains original value of b

A similar effect can be had by using the exclusive-or operator. This fact is widely used in computer graphics. A region of the computer screen can be highlighted by XORing the outline of a box around it. XORing the box outline a second time restores the original contents of the screen.

Doubly Linked List: A Complete Implementation Guide

Lesson 5 of 54By Simplilearn

Last updated on Sep 19, 20213097
Do doubly linked lists need a head?
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A Doubly linked list is used in navigation systems or to represent a classic deck of cards. A Doubly linked list is a bidirectional linked list; i.e., you can traverse it from head to tail node or tail to head node. Unlike singly-linked lists, its node has an extra pointer that points at the last node.

How Do You Implement a Doubly Linked List?

Do doubly linked lists need a head?

You create nodes of doubly-linked lists using classes or structures. These nodes are then linked with each other using the next and the previous pointer.

Code:

//A c++ program to implement linked list

#include <bits/stdc++.h>

using namespace std;

/* A class to create node */

class Node

{

public:

int data;

Node *next;

Node *prev;

};

//A function to insert at the

//beginning of the list

void push(Node** head, int newdata)

{

//create new node

Node* newnode = new Node();

/* put in the data */

newnode->data = newdata;

/* As we are adding at the beginning,

prev is always NULL */

newnode->prev = NULL;

/* link new node's next to head */

newnode->next = (*head);

/* change prev of head node to newnode */

if((*head) != NULL)

(*head)->prev = newnode ;

/* changing head node */

(*head) = newnode;

}

/* A c++ program to print the list */

void printlist(Node *head)

{

while(head != NULL)

{

cout << head->data << " ";

head = head->next;

}

}

int main()

{

/* We will start with an empty list */

Node* head = NULL;

/*lets create a linked list: 2->3->5->7 */

push(&head, 7);

push(&head, 5);

push(&head, 3);

push(&head, 2);

cout << "Created Doubly Linked list:" << endl;

printlist(head);

return 0;

}

Do doubly linked lists need a head?

Doubly Linked Lists - Insertion

Insertion in doubly linked lists is similar to what we saw in the singly linked list with two exceptions:

  1. We must update both the previous and next pointers in all affected nodes.
  2. We can use the tail pointer to make the insertion of data at the end of the list very efficient.