Print reverse of a Linked List without actually reversing in C language

CServer Side ProgrammingProgramming

The task is to print the reverse of a given linked list by using recursive function. The program must print the reverse but not reverse the list that means the order of the nodes remains the same

Here, the program will move head pointer containing the address of a first node to next node until the NULL is examined which is stored in last node of a list and printing the data of a head node.


Input: 29 34 43 56
Output: 56 43 34 29

Firstly, the nodes are inserted into the list and a pointer will start pointing to the nodes that are inserted. After final list is created let’s say temp pointer will be initialised with the first node pointer and it keeps incrementing till the node’s next address is NULL as the last node points to nothing and from the last node the list is traversed till the head pointer. It will display the reverse of a list without actually reversing a list.

The below code shows the c implementation of the algorithm given.


   Step 1 -> create node variable of type structure
      Declare int data
      Declare pointer of type node using *next
   Step 2 ->Declare function void reverse(node* head)
      IF head == NULL
      Call reverse(head->next)
      Print head->data
   Step 3 -> Declare Function void push(node** header, char newdata)
      Allocate memory using malloc
      Set newnode->data = newdata
      Set newnode->next = (*header)
      Set (*header) = newnode
   Step 4 ->In Main()
      Create list using node* head = NULL
      Insert elements through push(&head, 56)
      Call reverse(head)


//creating structure for a node
struct node {
   int data;
   node* next;
//function to print reverse of the data in the list
void reverse(node* head) {
   if (head == NULL)
   printf("%d ", head->data);
//function to puch the node in the list
void push(node** header, char newdata) {
   struct node* newnode = (struct node*)malloc(sizeof(struct node));
   newnode->data = newdata;
   newnode->next = (*header);
   (*header) = newnode;
int main() {
   node* head = NULL;
   push(&head, 56); //calling function to push 56 in the list
   push(&head, 43);
   push(&head, 34);
   push(&head, 29);
   return 0;


If we run above program then it will generate following output.

reverse of a linked list 56 43 34 29
Published on 22-Aug-2019 08:48:12