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 listclass 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 codellist = 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 listusing 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 listvar 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 methodimport 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 codellist = 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 listusing 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 methodvar 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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