An interesting method to print reverse of a linked list


We are given a linked list, we need to print the linked list in reverse order.


Input : list : 5-> 15-> 20-> 25 
Output : Reversed Linked list : 25-> 20-> 15-> 5

Input : list : 85-> 15-> 4-> 20 
Output : Reversed Linked list : 20-> 4-> 15-> 85

Input : list : 85
Output : Reversed Linked list : 85

For printing a list in reverse order, we have already discussed Iterative and Recursive Methods to Reverse.

In this post, an interesting method is discussed, that doesn’t require recursion and does no modifications to list. The function also visits every node of linked list only once.

Trick : Idea behind printing a list in reverse order without any recursive function or loop is to use Carriage return (“r”). For this, we should have knowledge of length of list. Now, we should print n-1 blanck space and then print 1st element then “r”, futher again n-2 blank space and 2nd node then “r” and so on..
Carriage return (“r”) : It commands a printer (cursor or the display of a system console), to move the position of the cursor to the first position on the same line.

// C program to print reverse of 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_ref, int n)
    int j = 0;
    struct Node* current = *head_ref;
    while (current != NULL) {

        // For each node, print proper number
        // of spaces before printing it
        for (int i = 0; i < 2 * (n - j); i++)
            printf(" ");

        // use of carriage return to move back
        // and print.
        printf("%dr", current->data);

        current = current->next;

/* Function to push a node */
void push(struct Node** head_ref, int new_data)
    struct Node* new_node = 
          (struct Node*)malloc(sizeof(struct Node));

    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;

/* Function to print linked list and find its
   length */
int printList(struct Node* head)
    // i for finding length of list
    int i = 0;
    struct Node* temp = head;
    while (temp != NULL) {
        printf("%d  ", temp->data);
        temp = temp->next;
    return i;

/* Driver program to test above function*/
int main()
    /* Start with the empty list */
    struct Node* head = NULL;
    // list nodes are as 6 5 4 3 2 1
    push(&head, 1);
    push(&head, 2);
    push(&head, 3);
    push(&head, 4);
    push(&head, 5);
    push(&head, 6);

    printf("Given linked list:n");
    // printlist print the list and
    // return the size of list
    int n = printList(head);

    // print reverse list with help
    // of carriage return function
    printf("nReversed Linked list:n");
    printReverse(&head, n);

    return 0;


Given linked list:
6 5 4 3 2 1
Reversed Linked List:
1 2 3 4 5 6

Input and Output Illustration :

NOTE:Above program may not work on online compiler because they do not support anything like carriage return on their console.

Reference :

stackoverflow/Carriage return

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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