Recursively Reversing a linked list (A simple implementation)

Given pointer to the head node of a linked list, the task is to recursively reverse the linked list. We need to reverse the list by changing links between nodes. 

Examples: 

Input : Head of following linked list  
       1->2->3->4->NULL
Output : Linked list should be changed to,
       4->3->2->1->NULL

Input : Head of following linked list  
       1->2->3->4->5->NULL
Output : Linked list should be changed to,
       5->4->3->2->1->NULL

Input : NULL
Output : NULL

Input  : 1->NULL
Output : 1->NULL

We have discussed an iterative and two recursive approaches in previous post on reverse a linked list. In this approach of reversing a linked list by passing a single pointer what we are trying to do is that we are making the previous node of the current node as his next node to reverse the linked list.

1. We return the pointer of next node to his previous(current) node and then make the previous node as the next node of returned node and then return the current node.
2. We first traverse to the last node and make the last node the head node of reversed linked list and then apply the above procedure in a recursive manner.

Implementation:

C++

// Recursive C++ program to reverse
// a linked list
#include <iostream>
using namespace std;

/* Link list node */
struct Node {
    int data;
    struct Node* next;
    Node(int data)
    {
        this->data = data;
        next = NULL;
    }
};

struct LinkedList {
    Node* head;
    LinkedList()
    {
        head = NULL;
    }

    /* Function to reverse the linked list */
    Node* reverse(Node* node)
    {
        if (node == NULL)
            return NULL;
        if (node->next == NULL) {
            head = node;
            return node;
        }
        Node* node1 = reverse(node->next);
        node1->next = node;
        node->next = NULL;
        return node;
    }

    /* Function to print linked list */
    void print()
    {
        struct Node* temp = head;
        while (temp != NULL) {
            cout << temp->data << " ";
            temp = temp->next;
        }
    }

    void push(int data)
    {
        Node* temp = new Node(data);
        temp->next = head;
        head = temp;
    }
};

/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    LinkedList ll;
    ll.push(20);
    ll.push(4);
    ll.push(15);
    ll.push(85);

    cout << "Given linked list\n";
    ll.print();

    ll.reverse(ll.head);

    cout << "\nReversed Linked list \n";
    ll.print();
    return 0;
}

Java

// Recursive Java program to reverse
// a linked list
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.util.Scanner;

public class ReverseLinkedListRecursive {

    /* Link list node */
    static class Node {
        public int data;
        public Node next;

        public Node(int nodeData) {
            this.data = nodeData;
            this.next = null;
        }
    }

    static class LinkedList {
        public Node head;

        public LinkedList() {
            this.head = null;
        }

        public void insertNode(int nodeData) {
            Node node = new Node(nodeData);

            if (this.head != null) {
                node.next = head;
            }
            this.head = node;
        }
    }

    /* Function to print linked list */
    public static void printSinglyLinkedList(Node node,
                        String sep) throws IOException {
        while (node != null) {
            System.out.print(String.valueOf(node.data) + sep);
            node = node.next;
        }
    }

    // Complete the reverse function below.
    static Node reverse(Node head) {
        if(head == null) {
            return head;
        }

        // last node or only one node
        if(head.next == null) {
            return head;
        }

        Node newHeadNode = reverse(head.next);

        // change references for middle chain
        head.next.next = head;
        head.next = null;

        // send back new head node in every recursion
        return newHeadNode;
    }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
            LinkedList llist = new LinkedList();

            llist.insertNode(20);
            llist.insertNode(4);
            llist.insertNode(15);
            llist.insertNode(85);

            System.out.println("Given linked list:");
            printSinglyLinkedList(llist.head, " ");

            System.out.println();
            System.out.println("Reversed Linked list:");
            Node llist1 = reverse(llist.head);
            printSinglyLinkedList(llist1, " ");

        scanner.close();
    }
}

Python3

# Recursive Python3 program to reverse
# a linked list
import math

# Link list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def LinkedList():
    head = None

# Function to reverse the linked list
def reverse(node):
    if (node == None):
        return node

    if (node.next == None):
        return node

    node1 = reverse(node.next)
    node.next.next = node
    node.next = None
    return node1

# Function to print linked list
def printList():
    temp = head
    while (temp != None) :
        print(temp.data, end = " ")
        temp = temp.next

def push(head_ref, new_data):
    new_node = Node(new_data)
    new_node.data = new_data
    new_node.next = head_ref
    head_ref = new_node
    return head_ref

# Driver Code
if __name__=='__main__':

    # Start with the empty list
    head = LinkedList()
    head = push(head, 20)
    head = push(head, 4)
    head = push(head, 15)
    head = push(head, 85)

    print("Given linked list")
    printList()

    head = reverse(head)

    print("\nReversed Linked list")
    printList()

# This code is contributed by AbhiThakur

C#

// Recursive C# program to reverse
// a linked list
using System;

public class ReverseLinkedListRecursive
{

    /* Link list node */
    public class Node
    {
        public int data;
        public Node next;

        public Node(int nodeData)
        {
            this.data = nodeData;
            this.next = null;
        }
    }

    class LinkedList
    {
        public Node head;

        public LinkedList()
        {
            this.head = null;
        }

        public void insertNode(int nodeData)
        {
            Node node = new Node(nodeData);

            if (this.head != null)
            {
                node.next = head;
            }
            this.head = node;
        }
    }

    /* Function to print linked list */
    public static void printSinglyLinkedList(Node node,
                        String sep)
    {
        while (node != null)
        {
            Console.Write(node.data + sep);
            node = node.next;
        }
    }

    // Complete the reverse function below.
    static Node reverse(Node head)
    {
        if(head == null)
        {
            return head;
        }

        // last node or only one node
        if(head.next == null)
        {
            return head;
        }

        Node newHeadNode = reverse(head.next);

        // change references for middle chain
        head.next.next = head;
        head.next = null;

        // send back new head
        // node in every recursion
        return newHeadNode;
    }

    // Driver code
    public static void Main(String[] args)
    {
            LinkedList llist = new LinkedList();

            llist.insertNode(20);
            llist.insertNode(4);
            llist.insertNode(15);
            llist.insertNode(85);

            Console.WriteLine("Given linked list:");
            printSinglyLinkedList(llist.head, " ");

            Console.WriteLine();
            Console.WriteLine("Reversed Linked list:");
            Node llist1 = reverse(llist.head);
            printSinglyLinkedList(llist1, " ");
    }
}

// This code has been contributed by Rajput-Ji

Javascript

// Recursive JavaScirpt program to reverse
// a linked list

// Link List node
class Node{
    constructor(data){
        this.data = data;
        this.next = null;
    }
}
let head = null;
// Function to reverse the linked list
function reverse(node){
    if(node == null) return null;

    if(node.next == null){
        head = node;
        return node;
    }

    let node1 = reverse(node.next);
    node1.next = node;
    node.next = null;
    return node;
}

// Function to print linked list
function print(){
    let temp = head;
    while(temp != null){
        console.log(temp.data + " ");
        temp = temp.next;
    }
}

function push(data){
    let temp = new Node(data);
    temp.next = head;
    head = temp;
}

// Driver Code
push(20);
push(4);
push(15);
push(85);

console.log("Given Linked List ");
print();

reverse(head)

console.log("\nReversed Linked List");
print();

// This code is contributed by Yash Agarwal(yashagarwal2852002)

Output

Given linked list
85 15 4 20 
Reversed Linked list 
20 4 15 85 

Time Complexity: O(N) where N is the number of elements in the linked list.
Auxiliary Space: O(N) due to stack recursion call.