Reverse a Linked List

Last Updated : 27 Feb, 2026

Given the head of a linked list, reverse the list and return the new head.

Examples: 

Input:

2

Output: 5 -> 4 -> 3 -> 2 -> 1

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Table of Content

[Approach] Using Iterative Method - O(n) Time and O(1) Space

The idea is to reverse the linked list by changing the direction of links using three pointers: prev, curr, and next. At each step, point the current node to its previous node and then move all three pointers forward until the list is fully reversed.

C++ 
#include <iostream>
using namespace std;

class Node {
  public:
    int data;
    Node *next;

    Node(int new_data) {
        data = new_data;
        next = nullptr;
    }
};

Node *reverseList(Node *head) {

    Node *curr = head, *prev = nullptr, *next;

    // Traverse all the nodes of Linked List
    while (curr != nullptr) {

        // Store next
        next = curr->next;

        // Reverse current node's next pointer
        curr->next = prev;

        // Move pointers one position ahead
        prev = curr;
        curr = next;
    }

    return prev;
}

void printList(Node *node) {
    while (node != nullptr) {
        cout << node->data;
        if (node->next)
            cout << " -> ";
        node = node->next;
    }
}

int main() {

    Node *head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);
    head->next->next->next->next = new Node(5);

    head = reverseList(head);

    printList(head);

    return 0;
}
C 
#include <stdio.h>

struct Node {
    int data;
    struct Node *next;
};

struct Node *reverseList(struct Node *head) {

    struct Node *curr = head, *prev = NULL, *next;

    // traverse all the nodes of Linked List
    while (curr != NULL) {

        // store next
        next = curr->next;

        // reverse current node's next pointer
        curr->next = prev;

        // move pointers one position ahead
        prev = curr;
        curr = next;
    }

    return prev;
}

void printList(struct Node *node) {
    while (node != NULL) {
        printf("%d", node->data);
        if (node->next)
            printf(" -> ");
        node = node->next;
    }
}

struct Node *createNode(int new_data) {
    struct Node *new_node = (struct Node *)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = NULL;
    return new_node;
}

int main() {

    struct Node *head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);
    head->next->next->next = createNode(4);
    head->next->next->next->next = createNode(5);

    head = reverseList(head);

    printList(head);

    return 0;
}
Java 
class Node {
    int data;
    Node next;

    Node(int new_data) {
        data = new_data;
        next = null;
    }
}

class GfG {
    static Node reverseList(Node head) {

        Node curr = head, prev = null, next;

        // Traverse all the nodes of Linked List
        while (curr != null) {

            // Store next
            next = curr.next;

            // Reverse current node's next pointer
            curr.next = prev;

            // Move pointers one position ahead
            prev = curr;
            curr = next;
        }

        return prev;
    }

    static void printList(Node node) {
        while (node != null) {
            System.out.print(node.data);
            if (node.next != null)
                System.out.print(" -> ");
            node = node.next;
        }
    }

    public static void main(String[] args){

        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);

        head = reverseList(head);
        printList(head);
    }
}
Python 
class Node:
    def __init__(self, newData):
        self.data = newData
        self.next = None

def reverseList(head):

    curr = head
    prev = None

    # traverse all the nodes of Linked List
    while curr is not None:

        # store next
        nextNode = curr.next

        # reverse current node's next pointer
        curr.next = prev

        # move pointers one position ahead
        prev = curr
        curr = nextNode

    return prev


def printList(node):
    while node is not None:
        print(f"{node.data}", end="")
        if node.next is not None:
            print(" -> ", end="")
        node = node.next
    print()


if __name__ == "__main__":

    head = Node(1)
    head.next = Node(2)
    head.next.next = Node(3)
    head.next.next.next = Node(4)
    head.next.next.next.next = Node(5)
    head = reverseList(head)
    printList(head)
C# 
using System;

class Node {
    public int Data;
    public Node Next;

    public Node(int newData)
    {
        Data = newData;
        Next = null;
    }
}

class GfG {
    static Node ReverseList(Node head) {

        Node curr = head;
        Node prev = null;
        Node next;

        // traverse all the nodes of Linked List
        while (curr != null) {

            // store next
            next = curr.Next;

            // reverse current node's next pointer
            curr.Next = prev;

            // move pointers one position ahead
            prev = curr;
            curr = next;
        }
        
        return prev;
    }

    static void printList(Node node) {
        while (node != null) {
            Console.Write("" + node.Data);
            if (node.Next != null)
                Console.Write(" -> ");
            node = node.Next;
        }
        Console.WriteLine();
    }

    static void Main() {

        Node head = new Node(1);
        head.Next = new Node(2);
        head.Next.Next = new Node(3);
        head.Next.Next.Next = new Node(4);
        head.Next.Next.Next.Next = new Node(5);
        head = ReverseList(head);
        printList(head);
    }
}
JavaScript 
class Node {
    constructor(newData) {
        this.data = newData;
        this.next = null;
    }
}
 
function reverseList(head) {
 
    let curr = head;
    let prev = null;
    let next;
 
    // traverse all the nodes of Linked List
    while (curr !== null) {
        // store next
        next = curr.next;
 
        // reverse current node's next pointer
        curr.next = prev;
 
        // move pointers one position ahead
        prev = curr;
        curr = next;
    }
 
    return prev;
}
 
function printList(node) {
    let result = [];
    while (node !== null) {
        result.push(node.data);
        node = node.next;
    }
    console.log(result.join(" -> "));
}
 
// Driver Code

let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
 
head = reverseList(head);
printList(head);

Output
5 -> 4 -> 3 -> 2 -> 1

[Alternate Approach 1] Using Recursion Method- O(n) Time and O(n) Space

The idea is to use recursion to reach the last node of the list, which becomes the new head after reversal. As the recursion starts returning, each node makes its next node point back to itself, effectively reversing the links one by one until the entire list is reversed.

C++ 
#include <iostream>
using namespace std;

class Node {
  public:
    int data;
    Node *next;

    Node(int new_data) {
        data = new_data;
        next = nullptr;
    }
};

Node *reverseList(Node *head) {
    if (head == NULL || head->next == NULL)
        return head;

    // reverse the rest of linked list and put
    // the first element at the end
    Node *rest = reverseList(head->next);

    // Make the current head as last node of
    // remaining linked list
    head->next->next = head;

    // Update next of current head to NULL
    head->next = NULL;

    return rest;
}

void printList(Node *node) {
    while (node != nullptr) {
        cout << node->data;
        if (node->next)
            cout << " -> ";
        node = node->next;
    }
}

int main() {

    Node *head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);
    head->next->next->next->next = new Node(5);

    head = reverseList(head);
    printList(head);

    return 0;
}
C 
#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node *next;
};

struct Node *reverseList(struct Node *head) {
    if (head == NULL || head->next == NULL)
        return head;

    // reverse the rest of linked list and put
    // the first element at the end
    struct Node *rest = reverseList(head->next);

    // make the current head as last node of
    // remaining linked list
    head->next->next = head;

    // update next of current head to NULL
    head->next = NULL;

    return rest;
}

void printList(struct Node *node) {
    while (node != NULL) {
        printf("%d", node->data);
        if (node->next)
            printf(" -> ");
        node = node->next;
    }
    printf("\n");
}

struct Node *createNode(int new_data) {
    struct Node *new_node = (struct Node *)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = NULL;
    return new_node;
}

int main() {
    struct Node *head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);
    head->next->next->next = createNode(4);
    head->next->next->next->next = createNode(5);

    head = reverseList(head);
    printList(head);

    return 0;
}
Java 
class Node {
    int data;
    Node next;

    Node(int new_data) {
        data = new_data;
        next = null;
    }
}

class GfG {
    static Node reverseList(Node head) {
        
        if (head == null || head.next == null)
            return head;

        // reverse the rest of linked list and put
        // the first element at the end
        Node rest = reverseList(head.next);

        // make the current head as last node of
        // remaining linked list
        head.next.next = head;

        // update next of current head to null
        head.next = null;

        return rest;
    }

    static void printList(Node node) {
        while (node != null) {
            System.out.print(node.data);
            if (node.next != null)
                System.out.print(" -> ");
            node = node.next;
        }
    }

    public static void main(String[] args) {

        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);

        head = reverseList(head);
        printList(head);
    }
}
Python 
class Node:
    def __init__(self, newData):
        self.data = newData
        self.next = None

def reverseList(head):
    if head is None or head.next is None:
        return head

    # reverse the rest of linked list and put the
    # first element at the end
    rest = reverseList(head.next)

    # make the current head as last node of
    # remaining linked list
    head.next.next = head

    # update next of current head to NULL
    head.next = None

    return rest


def printList(node):
    while node is not None:
        print(f"{node.data}", end="")
        if node.next is not None:
            print(" -> ", end="")
        node = node.next
    print()


if __name__ == "__main__":

    # Create a hard-coded linked list:
    # 1 -> 2 -> 3 -> 4 -> 5
    head = Node(1)
    head.next = Node(2)
    head.next.next = Node(3)
    head.next.next.next = Node(4)
    head.next.next.next.next = Node(5)

    head = reverseList(head)
    printList(head)
C# 
using System;

class Node {
    public int Data;
    public Node Next;

    public Node(int newData) {
        Data = newData;
        Next = null;
    }
}

class GfG {
    static Node ReverseList(Node head) {
        
        if (head == null || head.Next == null)
            return head;

        // reverse the rest of linked list and
        // put the first element at the end
        Node rest = ReverseList(head.Next);

        // make the current head as last node
        // of remaining linked list
        head.Next.Next = head;

        // update next of current head to null
        head.Next = null;

        return rest;
    }

    static void PrintList(Node node) {
        while (node != null) {
            Console.Write("" + node.Data);
            if (node.Next != null)
                Console.Write(" -> ");
            node = node.Next;
        }
        Console.WriteLine();
    }

    static void Main() {
        Node head = new Node(1);
        head.Next = new Node(2);
        head.Next.Next = new Node(3);
        head.Next.Next.Next = new Node(4);
        head.Next.Next.Next.Next = new Node(5);

        head = ReverseList(head);
        PrintList(head);
    }
}
JavaScript 
class Node {
    constructor(new_data) {
        this.data = new_data;
        this.next = null;
    }
}
 
function reverseList(head) {
    if (head === null || head.next === null)
        return head;
 
    // reverse the rest of linked list
    let rest = reverseList(head.next);
 
    // make the current head as last node of
    // remaining linked list
    head.next.next = head;
 
    // update next of current head to null
    head.next = null;
 
    return rest;
}
 
function printList(node) {
    let result = [];
    while (node !== null) {
        result.push(node.data);
        node = node.next;
    }
    console.log(result.join(" -> "));
}
 
// Driver code
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
 
head = reverseList(head);
printList(head);

Output
5 -> 4 -> 3 -> 2 -> 1

[Alternate Approach 2] Using Stack - O(n) Time and O(n) Space

The idea is to traverse the linked list and push all nodes except the last node into the stack. Make the last node as the new headof the reversed linked list. Now, start popping the element and append each node to the reversed Linked List. Finally, return the head of the reversed linked list.

C++ 
#include <iostream>
#include <stack>
using namespace std;

class Node {
  public:
    int data;
    Node *next;

    Node(int new_data) {
        data = new_data;
        next = nullptr;
    }
};

Node *reverseList(Node *head) {
    
    stack<Node *> s;
    Node *temp = head;

    // push all nodes except the last node into stack
    while (temp->next != NULL) {
        s.push(temp);
        temp = temp->next;
    }

    // make the last node as new head of the linked list
    head = temp;

    // pop all the nodes and append to the linked list
    while (!s.empty()) {
        
        // append the top value of stack in list
        temp->next = s.top();

        s.pop();

        temp = temp->next;
    }

    // update the next pointer of last node of stack to null
    temp->next = NULL;

    return head;
}

void printList(Node *node) {
    while (node != nullptr) {
        cout << node->data;
        if (node->next)
            cout << " -> ";
        node = node->next;
    }
}

int main() {
    Node *head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);
    head->next->next->next->next = new Node(5);

    head = reverseList(head);
    printList(head);

    return 0;
}
C 
#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node *next;
};

struct Node *createNode(int new_data) {
    struct Node *new_node = (struct Node *)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = NULL;
    return new_node;
}

struct Node *reverseList(struct Node *head) {
    
    struct Node *stack[100000];
    int top = -1;
    struct Node *temp = head;

    // push all nodes into stack
    while (temp != NULL) {
        stack[++top] = temp;
        temp = temp->next;
    }

    // make the last node as new head of the linked list
    if (top >= 0) {
        head = stack[top];
        temp = head;

        // pop all the nodes and append to the linked list
        while (top > 0) {
            
            // append the top value of stack in list
            temp->next = stack[--top];

            temp = temp->next;
        }

        // update the next pointer of last node of stack to null
        temp->next = NULL;
    }

    return head;
}

void printList(struct Node *node) {
    while (node != NULL) {
        printf("%d", node->data);
        if (node->next)
            printf(" -> ");
        node = node->next;
    }
    printf("\n");
}

int main() {
    struct Node *head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);
    head->next->next->next = createNode(4);
    head->next->next->next->next = createNode(5);

    head = reverseList(head);
    printList(head);

    return 0;
}
Java 
import java.util.Stack;

class Node {
    int data;
    Node next;

    Node(int new_data) {
        data = new_data;
        next = null;
    }
}

class GfG {
    static Node reverseList(Node head) {
        Stack<Node> stack = new Stack<>();
        Node temp = head;

        // push all nodes into stack
        while (temp != null) {
            stack.push(temp);
            temp = temp.next;
        }

        // make the last node as new head of the linked list
        if (!stack.isEmpty()) {
            head = stack.pop();
            temp = head;

            // pop all the nodes and append to the linked list
            while (!stack.isEmpty()) {
                
                // append the top value of stack in list
                temp.next = stack.pop();

                temp = temp.next;
            }

            // update the next pointer of last node to null
            temp.next = null;
        }

        return head;
    }

    static void printList(Node node) {
        while (node != null) {
            System.out.print(node.data);
            if (node.next != null)
                System.out.print(" -> ");
            node = node.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);

        head = reverseList(head);
        printList(head);
    }
}
Python 
class Node:
    def __init__(self, new_data):

        self.data = new_data
        self.next = None


def reverseList(head):

    stack = []

    temp = head

    # push all nodes except the last node into stack
    while temp.next is not None:
        stack.append(temp)
        temp = temp.next

    head = temp

    # pop all the nodes and append to the linked list
    while stack:

        # append the top value of stack in list
        temp.next = stack.pop()

        temp = temp.next

    temp.next = None

    return head


def printList(node):
    while node is not None:
        print(f"{node.data}", end="")
        if node.next is not None:
            print(" -> ", end="")
        node = node.next
    print()

if __name__ == "__main__":
    # create a hard-coded linked list:
    # 1 -> 2 -> 3 -> 4 -> 5
    head = Node(1)
    head.next = Node(2)
    head.next.next = Node(3)
    head.next.next.next = Node(4)
    head.next.next.next.next = Node(5)

    head = reverseList(head)
    printList(head)
C# 
using System;
using System.Collections.Generic;

class Node {
    public int Data;
    public Node Next;

    public Node(int newData) {
        Data = newData;
        Next = null;
    }
}

class GfG {
    static Node ReverseList(Node head) {
        
        Stack<Node> stack = new Stack<Node>();
        Node temp = head;

        // push all nodes into stack
        while (temp != null) {
            stack.Push(temp);
            temp = temp.Next;
        }

        // make the last node as new head of the linked list
        if (stack.Count > 0) {
            head = stack.Pop();
            temp = head;

            // pop all the nodes and append to the linked list
            while (stack.Count > 0) {
                
                temp.Next = stack.Pop();

                temp = temp.Next;
            }

            temp.Next = null;
        }

        return head;
    }

    static void printList(Node node) {
        while (node != null) {
            Console.Write("" + node.Data);
            if (node.Next != null)
                Console.Write(" -> ");
            node = node.Next;
        }
        Console.WriteLine();
    }

    static void Main() {
        Node head = new Node(1);
        head.Next = new Node(2);
        head.Next.Next = new Node(3);
        head.Next.Next.Next = new Node(4);
        head.Next.Next.Next.Next = new Node(5);

        head = ReverseList(head);
        printList(head);
    }
}
JavaScript 
class Node {
    constructor(newData) {
        this.data = newData;
        this.next = null;
    }
}
 
function reverseList(head) {
    let stack = [];
    let temp = head;
 
    // push all nodes into stack
    while (temp !== null) {
        stack.push(temp);
        temp = temp.next;
    }
 
    // make the last node as new head of the linked list
    if (stack.length > 0) {
        head = stack.pop();
        temp = head;
 
        // pop all the nodes and append to the linked list
        while (stack.length > 0) {
            
            temp.next = stack.pop();
 
            temp = temp.next;
        }
 
        temp.next = null;
    }
 
    return head;
}
 
function printList(node) {
    let result = [];
    while (node !== null) {
        result.push(node.data);
        node = node.next;
    }
    console.log(result.join(" -> "));
}
 
// driver code
 
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);

head = reverseList(head);
printList(head);

Output
5 -> 4 -> 3 -> 2 -> 1
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