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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
