Reverse a Doubly Linked List without swapping nodes

Write a program to reverse the given Doubly Linked List.

See below diagrams for example.

          (a) Original Doubly Linked List  

          (b) Reversed Doubly Linked List  

Approach: In the previous post, doubly linked list is being reversed by swapping prev and next pointers for all nodes, changing prev of the head (or start) and then changing the head pointer in the end. In this post, we create a push function that adds the given node at the beginning of the given list. We traverse the original list and one by one pass the current node pointer to the push function. This process will reverse the list. Finally return the new head of this reversed list.

C++

// C++ implementation to reverse
// a doubly linked list
#include <bits/stdc++.h>

using namespace std;

// a node of the doubly linked list
struct Node {
    int data;
    Node *next, *prev;
};

// function to get a new node
Node* getNode(int data)
{
    // allocate space
    Node* new_node = (Node*)malloc(sizeof(Node));

    // put in the data
    new_node->data = data;
    new_node->next = new_node->prev = NULL;
    return new_node;
}

// function to insert a node at the beginning
// of the Doubly Linked List
void push(Node** head_ref, Node* new_node)
{
    // since we are adding at the beginning,
    // prev is always NULL
    new_node->prev = NULL;

    // link the old list of the new node
    new_node->next = (*head_ref);

    // change prev of head node to new node
    if ((*head_ref) != NULL)
        (*head_ref)->prev = new_node;

    // move the head to point to the new node
    (*head_ref) = new_node;
}

// function to reverse a doubly linked list
void reverseList(Node** head_ref)
{
    // if list is empty or it contains
    // a single node only
    if (!(*head_ref) || !((*head_ref)->next))
        return;

    Node* new_head = NULL;
    Node *curr = *head_ref, *next;

    while (curr != NULL) {

        // get pointer to next node
        next = curr->next;

        // push 'curr' node at the beginning of the
        // list with starting with 'new_head'
        push(&new_head, curr);

        // update 'curr'
        curr = next;
    }

    // update 'head_ref'
    *head_ref = new_head;
}

// Function to print nodes in a
// given doubly linked list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}

// Driver program to test above
int main()
{
    // Start with the empty list
    Node* head = NULL;

    // Create doubly linked: 10<->8<->4<->2 */
    push(&head, getNode(2));
    push(&head, getNode(4));
    push(&head, getNode(8));
    push(&head, getNode(10));

    cout << "Original list: ";
    printList(head);

    // Reverse doubly linked list
    reverseList(&head);

    cout << "\nReversed list: ";
    printList(head);

    return 0;
}

Java

// Java implementation to reverse
// a doubly linked list
class GFG
{

// a node of the doubly linked list
static class Node
{
    int data;
    Node next, prev;
};

// function to get a new node
static Node getNode(int data)
{
    // allocate space
    Node new_node = new Node();

    // put in the data
    new_node.data = data;
    new_node.next = new_node.prev = null;
    return new_node;
}

// function to insert a node at the beginning
// of the Doubly Linked List
static Node push(Node head_ref, Node new_node)
{
    // since we are adding at the beginning,
    // prev is always null
    new_node.prev = null;

    // link the old list of the new node
    new_node.next = (head_ref);

    // change prev of head node to new node
    if ((head_ref) != null)
        (head_ref).prev = new_node;

    // move the head to point to the new node
    (head_ref) = new_node;

    return head_ref;
}

// function to reverse a doubly linked list
static Node reverseList(Node head_ref)
{
    // if list is empty or it contains
    // a single node only
    if ((head_ref) == null || ((head_ref).next) == null)
        return null;

    Node new_head = null;
    Node curr = head_ref, next;

    while (curr != null)
    {

        // get pointer to next node
        next = curr.next;

        // push 'curr' node at the beginning of the
        // list with starting with 'new_head'
        new_head = push(new_head, curr);

        // update 'curr'
        curr = next;
    }

    // update 'head_ref'
    head_ref = new_head;

    return head_ref;
}

// Function to print nodes in a
// given doubly linked list
static void printList(Node head)
{
    while (head != null)
    {
        System.out.print(head.data + " ");
        head = head.next;
    }
}

// Driver program to test above
public static void main(String args[])
{
    // Start with the empty list
    Node head = null;

    // Create doubly linked: 10< - >8< - >4< - >2 /
    head = push(head, getNode(2));
    head = push(head, getNode(4));
    head = push(head, getNode(8));
    head = push(head, getNode(10));

    System.out.print("Original list: ");
    printList(head);

    // Reverse doubly linked list
    head = reverseList(head);

    System.out.print("\nReversed list: ");
    printList(head);
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation to reverse
# a doubly linked list
import math

# a node of the doubly linked list
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

# function to get a new node
def getNode(data):

    # allocate space
    new_node = Node(data)

    # put in the data
    new_node.data = data
    new_node.next = None
    new_node.prev = None
    return new_node

# function to insert a node at the beginning
# of the Doubly Linked List
def push(head_ref, new_node):

    # since we are adding at the beginning,
    # prev is always None
    new_node.prev = None

    # link the old list of the new node
    new_node.next = head_ref

    # change prev of head node to new node
    if (head_ref != None):
        head_ref.prev = new_node

    # move the head to point to the new node
    head_ref = new_node
    return head_ref

# function to reverse a doubly linked list
def reverseList(head_ref):

    # if list is empty or it contains
    # a single node only
    if (head_ref == None or
       (head_ref).next == None):
        return None

    new_head = None
    curr = head_ref

    while (curr != None):

        # get pointer to next node
        next = curr.next

        # push 'curr' node at the beginning of the
        # list with starting with 'new_head'
        new_head = push(new_head, curr)

        # update 'curr'
        curr = next

    # update 'head_ref'
    head_ref = new_head

    return head_ref

# Function to print  nodes in a
# given doubly linked list
def printList(head):
    while (head != None) :
        print(head.data, end = " ")
        head = head.next

# Driver Code
if __name__=='__main__':

    # Start with the empty list
    head = None

    # Create doubly linked: 10<.8<.4<.2 */
    head = push(head, getNode(2));
    head = push(head, getNode(4));
    head = push(head, getNode(8));
    head = push(head, getNode(10));

    print("Original list: ", end = "")
    printList(head)

    # Reverse doubly linked list
    head = reverseList(head)

    print("\nReversed list: ", end = "")
    printList(head)

# This code is contributed by Srathore

C#

// C# implementation to reverse
// a doubly linked list
using System;

class GFG
{

// a node of the doubly linked list
public class Node
{
    public int data;
    public Node next, prev;
};

// function to get a new node
static Node getNode(int data)
{
    // allocate space
    Node new_node = new Node();

    // put in the data
    new_node.data = data;
    new_node.next = new_node.prev = null;
    return new_node;
}

// function to insert a node at the beginning
// of the Doubly Linked List
static Node push(Node head_ref, Node new_node)
{
    // since we are adding at the beginning,
    // prev is always null
    new_node.prev = null;

    // link the old list of the new node
    new_node.next = (head_ref);

    // change prev of head node to new node
    if ((head_ref) != null)
        (head_ref).prev = new_node;

    // move the head to point to the new node
    (head_ref) = new_node;

    return head_ref;
}

// function to reverse a doubly linked list
static Node reverseList(Node head_ref)
{
    // if list is empty or it contains
    // a single node only
    if ((head_ref) == null || ((head_ref).next) == null)
        return null;

    Node new_head = null;
    Node curr = head_ref, next;

    while (curr != null)
    {

        // get pointer to next node
        next = curr.next;

        // push 'curr' node at the beginning of the
        // list with starting with 'new_head'
        new_head = push(new_head, curr);

        // update 'curr'
        curr = next;
    }

    // update 'head_ref'
    head_ref = new_head;

    return head_ref;
}

// Function to print nodes in a
// given doubly linked list
static void printList(Node head)
{
    while (head != null)
    {
        Console.Write(head.data + " ");
        head = head.next;
    }
}

// Driver code
public static void Main(String []args)
{
    // Start with the empty list
    Node head = null;

    // Create doubly linked: 10< - >8< - >4< - >2 /
    head = push(head, getNode(2));
    head = push(head, getNode(4));
    head = push(head, getNode(8));
    head = push(head, getNode(10));

    Console.Write("Original list: ");
    printList(head);

    // Reverse doubly linked list
    head = reverseList(head);

    Console.Write("\nReversed list: ");
    printList(head);
}
}

// This code has been contributed by 29AjayKumar

Javascript

<script>
// javascript implementation to reverse
// a doubly linked list     // a node of the doubly linked list
class Node {
    constructor() {
        this.data = 0;
        this.prev = null;
        this.next = null;
    }
}
    // function to get a new node
    function getNode(data) {
        // allocate space
var new_node = new Node();

        // put in the data
        new_node.data = data;
        new_node.next = new_node.prev = null;
        return new_node;
    }

    // function to insert a node at the beginning
    // of the Doubly Linked List
    function push(head_ref,  new_node) {
        // since we are adding at the beginning,
        // prev is always null
        new_node.prev = null;

        // link the old list of the new node
        new_node.next = (head_ref);

        // change prev of head node to new node
        if ((head_ref) != null)
            (head_ref).prev = new_node;

        // move the head to point to the new node
        (head_ref) = new_node;

        return head_ref;
    }

    // function to reverse a doubly linked list
    function reverseList(head_ref) {
        // if list is empty or it contains
        // a single node only
        if ((head_ref) == null || ((head_ref).next) == null)
            return null;

var new_head = null;
var curr = head_ref, next;

        while (curr != null) {

            // get pointer to next node
            next = curr.next;

            // push 'curr' node at the beginning of the
            // list with starting with 'new_head'
            new_head = push(new_head, curr);

            // update 'curr'
            curr = next;
        }

        // update 'head_ref'
        head_ref = new_head;

        return head_ref;
    }

    // Function to print nodes in a
    // given doubly linked list
    function printList(head) {
        while (head != null) {
            document.write(head.data + " ");
            head = head.next;
        }
    }

    // Driver program to test above

        // Start with the empty list
var head = null;

        // Create doubly linked: 10< - >8< - >4< - >2 /
        head = push(head, getNode(2));
        head = push(head, getNode(4));
        head = push(head, getNode(8));
        head = push(head, getNode(10));

        document.write("Original list: ");
        printList(head);

        // Reverse doubly linked list
        head = reverseList(head);

        document.write("<br/>Reversed list: ");
        printList(head);

// This code contributed by Rajput-Ji
</script>

Output:

Original list: 10 8 4 2
Reversed list: 2 4 8 10

Time Complexity: O(n).
 Auxiliary Space: O(1)