Print reverse of a Linked List without actually reversing

Given a linked list, print reverse of it using a recursive function. For example, if the given linked list is 1->2->3->4, then output should be 4->3->2->1.
Note that the question is only about printing the reverse. To reverse the list itself see this
Difficulty Level: Rookie 

Algorithm 

printReverse(head)
  1. call print reverse for head->next
  2. print head->data

Implementation:

C++

// C++ program to print reverse of a linked list
#include <bits/stdc++.h>
using namespace std;

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

/* Function to reverse the linked list */
void printReverse(Node* head)
{
    // Base case
    if (head == NULL)
    return;

    // print the list after head node
    printReverse(head->next);

    // After everything else is printed, print head
    cout << head->data << " ";
}

/*UTILITY FUNCTIONS*/
/* Push a node to linked list. Note that this function
changes the head */
void push(Node** head_ref, char new_data)
{
    /* allocate node */
    Node* new_node = new Node();

    /* put in the data */
    new_node->data = new_data;

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

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

/* Driver code*/
int main()
{
    // Let us create linked list 1->2->3->4
    Node* head = NULL;
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);

    printReverse(head);
    return 0;
}

// This code is contributed by rathbhupendra

C

// C program to print reverse of a linked list
#include<stdio.h>
#include<stdlib.h>

/* Link list node */
struct Node
{
    int data;
    struct Node* next;
};

/* Function to reverse the linked list */
void printReverse(struct Node* head)
{
    // Base case
    if (head == NULL)
       return;

    // print the list after head node
    printReverse(head->next);

    // After everything else is printed, print head
    printf("%d  ", head->data);
}

/*UTILITY FUNCTIONS*/
/* Push a node to linked list. Note that this function
  changes the head */
void push(struct Node** head_ref, char new_data)
{
    /* allocate node */
    struct Node* new_node =
            (struct Node*) malloc(sizeof(struct Node));

    /* put in the data  */
    new_node->data  = new_data;

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

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

/* Driver program to test above function*/
int main()
{
    // Let us create linked list 1->2->3->4
    struct Node* head = NULL;
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);

    printReverse(head);
    return 0;
}

Java

// Java program to print reverse of a linked list
class LinkedList
{
    Node head;  // head of list

    /* Linked list Node*/
    class Node
    {
        int data;
        Node next;
        Node(int d) {data = d; next = null; }
    }

    /* Function to print reverse of linked list */
    void printReverse(Node head)
    {
        if (head == null) return;

        // print list of head node
        printReverse(head.next);

        // After everything else is printed
        System.out.print(head.data+" ");
    }

    /* Utility Functions */

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /*Driver function to test the above methods*/
    public static void main(String args[])
    {
        // Let us create linked list 1->2->3->4
        LinkedList llist = new LinkedList();
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);

        llist.printReverse(llist.head);
    }
}
/* This code is contributed by Rajat Mishra */

Python3

# Python3 program to print reverse
# of a linked list

# Node class
class Node:

    # Constructor to initialize
    # the node object
    def __init__(self, data):

        self.data = data
        self.next = None

class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    # Recursive function to print
    # linked list in reverse order
    def printrev(self, temp):

        if temp:
            self.printrev(temp.next)
            print(temp.data, end = ' ')
        else:
            return

    # Function to insert a new node
    # at the beginning
    def push(self, new_data):

        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

# Driver code
llist = LinkedList()

llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)

llist.printrev(llist.head)

# This code is contributed by Vinay Kumar(vinaykumar71)

C#

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

public class LinkedList
{
    Node head; // head of list

    /* Linked list Node*/
    class Node
    {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d; next = null;
        }
    }

    /* Function to print reverse of linked list */
    void printReverse(Node head)
    {
        if (head == null) return;

        // print list of head node
        printReverse(head.next);

        // After everything else is printed
        Console.Write(head.data + " ");
    }

    /* Utility Functions */

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /*Driver function to test the above methods*/
    public static void Main(String []args)
    {
        // Let us create linked list 1->2->3->4
        LinkedList llist = new LinkedList();
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);

        llist.printReverse(llist.head);
    }
}

// This code has been contributed by Rajput-Ji

Javascript

<script>

// Javascript program to print reverse of a linked list
var head; // head of list

    /* Linked list Node */
    class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }

    /* Function to print reverse of linked list */
    function printReverse( head) {
        if (head == null)
            return;

        // print list of head node
        printReverse(head.next);

        // After everything else is printed
        document.write(head.data + " ");
    }

    /* Utility Functions */

    /* Inserts a new Node at front of the list. */
    function push(new_data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
         new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Driver function to test the above methods */

        // Let us create linked list 1->2->3->4

        push(4);
        push(3);
        push(2);
        push(1);

        printReverse(head);

// This code contributed by Rajput-Ji

</script>

Output

4 3 2 1 

Time Complexity: O(n) 

Auxiliary Space: O(n) for stack space since using recursion

Another approach:

We can also perform the same action using a stack using iterative method.

Algorithm:

Store the values of the linked list in a stack.
Keep removing the elements from the stack and print them.

Implementation:

C++

// C++ program to print reverse of a linked list using iterative method
#include <bits/stdc++.h>
using namespace std;

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

/* Function to print the reverse of the linked list using iterative method */
void printReverse(Node* head)
{
    stack<int> st;
    Node *curr = head;
    while(curr!=NULL)
    {
      st.push(curr->data);
      curr = curr->next;
    }
      while(st.empty()==false)
    {
      cout << st.top()<<" -> ";
      st.pop();
    }
}

/*UTILITY FUNCTIONS*/
/* function to print the elements of the linked list*/
void printList(Node *head)
{
  Node *curr = head;
  while(curr!=NULL)
  {
    cout << curr->data << " -> ";
    curr = curr->next;
  }
  cout<<"\n";
}
/* Push a node to linked list. Note that this function
changes the head */
void push(Node** head_ref, char new_data)
{
    /* allocate node */
    Node* new_node = new Node();

    /* put in the data */
    new_node->data = new_data;

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

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

/* Driver code*/
int main()
{
    // Let us create linked list 1->2->3->4
    Node* head = NULL;
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
    printList(head);
    printReverse(head);
    return 0;
}

// This code is contributed by Abhijeet Kumar(abhijeet19403)

Java

// Java program to print reverse of a linked list using iterative method
import java.util.*;
class LinkedList
{
    Node head;  // head of list

    /* Linked list Node*/
    class Node
    {
        int data;
        Node next;
        Node(int d) {data = d; next = null; }
    }

    /* Function to print reverse of linked list */
    void printReverse(Node head)
    {
        Stack<Integer> st = new Stack<Integer>();
        Node curr = head;
        while(curr!=null)
        {
          st.push(curr.data);
          curr = curr.next;
        }
          while(st.isEmpty()==false)
        {
          System.out.print(st.peek() + " -> ");
          st.pop();
        }
    }

    /* Utility Functions */
    /* function to print the elements of the linked list*/
    void printList(Node head)
    {
          Node curr = head;
          while(curr!=null)
          {
            System.out.print(curr.data + " -> ");
            curr = curr.next;
          }
         System.out.println();
    }
    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /*Driver function to test the above methods*/
    public static void main(String args[])
    {
        // Let us create linked list 1->2->3->4
        LinkedList llist = new LinkedList();
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);
        llist.printList(llist.head);
        llist.printReverse(llist.head);
    }
}
/* This code is contributed by Abhijeet Kumar(abhijeet19403) */

Python3

# Python3 program to print reverse
# of a linked list using iterative method

# Node class
class Node:

    # Constructor to initialize
    # the node object
    def __init__(self, data):

        self.data = data
        self.next = None

class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    # Recursive function to print
    # linked list in reverse order using iterative method
    def printrev(self, curr):
        stack = []
        while curr:
            stack.append(curr)
            curr = curr.next

        while len(stack):
            print(stack.pop().data,"->" , end = ' ')
        print()

    #Function to print the linked list
    def printlist(self,curr):
        while curr:
            print(curr.data,"->" , end = ' ')
            curr = curr.next
        print()

    # Function to insert a new node
    # at the beginning
    def push(self, new_data):

        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

# Driver code
llist = LinkedList()

llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)

llist.printlist(llist.head)
llist.printrev(llist.head)

# This code is contributed by Abhijeet Kumar(abhijeet19403)

C#

// C# program to print reverse of a linked list
using System;
using System.Collections;
public class LinkedList {
    Node head; // head of list

    /* Linked list Node*/
    class Node {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    /* Function to print reverse of linked list */
    void printReverse(Node head)
    {
        Stack st = new Stack();
        Node curr = head;
        while (curr != null) {
            st.Push(curr.data);
            curr = curr.next;
        }
        while (st.Count != 0) {
            Console.Write(st.Peek() + " -> ");
            st.Pop();
        }
        Console.Write("\n");
    }

    /* Utility Functions */
    /* A function to print the elements of the list*/
    void printList(Node head)
    {
        Node curr = head;
        while (curr != null) {
            Console.Write(curr.data + " -> ");
            curr = curr.next;
        }
        Console.Write("\n");
    }
    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /*Driver function to test the above methods*/
    public static void Main(String[] args)
    {
        // Let us create linked list 1->2->3->4
        LinkedList llist = new LinkedList();
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);
        llist.printList(llist.head);
        llist.printReverse(llist.head);
    }
}

// This code has been contributed by Abhijeet Kumar(abhijeet19403)

Javascript

<script>

// Javascript program to print reverse of a linked list using iterative method
var head; // head of list

    /* Linked list Node */
    class Node {
        constructor(val) {
            this.data = val;
            this.next = null;
        }
    }

    /* Function to print reverse of linked list using iterative method */
    function printReverse( head) {
        var st = new Stack();
        var curr = this.head;
        while(curr != null)
        {
            st.push(curr.data);
            curr = curr.next;
        }
        while(st.idEmpty())
        {
            document.write(head.data + " -> ");
            st.pop();
        }
    }

    /* Utility Functions */
    /* function to print the elements of the linked list*/
    void printList( head)
    {
          var curr = this.head;
          while(curr!=null)
          {
            document.write(curr.data + " -> ");
            curr = curr.next;
          }
         document.write("\n");
    }
    /* Inserts a new Node at front of the list. */
    function push(new_data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
         new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Driver function to test the above methods */

        // Let us create linked list 1->2->3->4

        push(4);
        push(3);
        push(2);
        push(1);
        printList(head);
        printReverse(head);

// This code contributed by Abhijeet Kumar(abhijeet19403)

</script>

Output

1 -> 2 -> 3 -> 4 -> 
4 -> 3 -> 2 -> 1 -> 

Time Complexity: O(N)

As we are traversing the linked list only once.

Auxiliary Space: O(N)

The extra space is used in storing the elements in the stack.

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