C Program to delete alternate nodes of a Linked List

September 18, 2025

Delete alternate nodes of a Linked List

C Program to delete alternate nodes of a Linked List. In this program, we want to remove alternate nodes from the singly linked list which is start from the second node of the linked list and remove all the alternate nodes of it.

For Example,

  • Input: 1 -> 3 -> 5 -> 7 -> 9
  • Output: 1 -> 5 -> 9 
delete alternate nodes

Working for deleting alternate nodes of a Singly Linked List:-

  • Take linked list input
  • Initialize prev and curr, assign prev = head & curr = head->next
  • Change next pointer of previous to curr->next, so link between prev & curr is broken
  • Now, curr is unreachable, free its memory
  • Move prev to next node in linked list
  • Move curr to next node.
  • Keep doing all above steps until you reach end of the node

Structure of creating a node of the singly linked list:-

struct node
{
    int data;
    struct node *next;
};
C Program to delete alternate nodes of a Linked List

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C Program to delete alternate nodes of a Linked List

Run 
#include<stdio.h>
#include<stdlib.h>

//Structure of node 
struct Node
{
  int data;
  struct Node *next;
};

//display the nodes
void display (struct Node *node)
{

  //as linked list will end when Node is Null
  while (node != NULL)
    {
      printf ("%d ", node->data);
      node = node->next;
    }
  printf ("\n");
}

// function to delete alternate nodes
void delete_Alt (struct Node *head)
{
  if (head == NULL)
    return;

  // prev req so its next node can be changed
  struct Node *prev = head;
  struct Node *curr = head->next;

  while (prev != NULL && curr != NULL)
    {
      // changing next of previous node
      prev->next = curr->next;

      // free the memory
      free (curr);

      // Update prev and curr 
      prev = prev->next;
      if (prev != NULL)
	curr = prev->next;
    }
}

int main ()
{
  //creating 4 pointers of type struct Node
  //So these can point to address of struct type variable
  struct Node *head = NULL;
  struct Node *node2 = NULL;
  struct Node *node3 = NULL;
  struct Node *node4 = NULL;
  struct Node *node5 = NULL;

  // allocate 4 nodes in the heap 
  head = (struct Node *) malloc (sizeof (struct Node));
  node2 = (struct Node *) malloc (sizeof (struct Node));
  node3 = (struct Node *) malloc (sizeof (struct Node));
  node4 = (struct Node *) malloc (sizeof (struct Node));
  node5 = (struct Node *) malloc (sizeof (struct Node));


  head->data = 5;		// data set for head node 
  head->next = node2;		// next pointer assigned to address of node2 

  node2->data = 4;
  node2->next = node3;

  node3->data = 3;
  node3->next = node4;

  node4->data = 2;
  node4->next = node5;

  node5->data = 1;
  node5->next = NULL;

  printf ("Before deletion: ");
  display (head);

  delete_Alt (head);

  printf ("After deletion: ");
  display (head);
  return 0;
}

Output

Before deletion: 5 4 3 2 1 
After deletion: 5 3 1
Above has a time complexity of O(N)

Now let us look at the recursive approach for the same.

Method 2 (using Recursion)

Below also has O(N) time complexity due to n order recursive calls.
Run 
#include<stdio.h>
#include<stdlib.h>

//Structure of node 
struct Node
{
  int data;
  struct Node *next;
};

//display the nodes
void display (struct Node *node)
{

  //as linked list will end when Node is Null
  while (node != NULL)
    {
      printf ("%d ", node->data);
      node = node->next;
    }
  printf ("\n");
}

// function to delete alternate nodes
void delete_Alt (struct Node *curr)
{
  if (curr == NULL)
    return;

  // assigning node to current node's next
  struct Node *del_node = curr->next;

  if (del_node == NULL)
    return;

  // changing next node of curr to alternative node
  // del_node is no more part of Linked List
  curr->next = del_node->next;

  // memory for del_node also freed
  free (del_node);

  // Recursively call for the new next of curr
  delete_Alt (curr->next);
}

int main ()
{
  //creating 4 pointers of type struct Node
  //So these can point to address of struct type variable
  struct Node *head = NULL;
  struct Node *node2 = NULL;
  struct Node *node3 = NULL;
  struct Node *node4 = NULL;
  struct Node *node5 = NULL;

  // allocate 4 nodes in the heap 
  head = (struct Node *) malloc (sizeof (struct Node));
  node2 = (struct Node *) malloc (sizeof (struct Node));
  node3 = (struct Node *) malloc (sizeof (struct Node));
  node4 = (struct Node *) malloc (sizeof (struct Node));
  node5 = (struct Node *) malloc (sizeof (struct Node));


  head->data = 5;		// data set for head node 
  head->next = node2;		// next pointer assigned to address of node2 

  node2->data = 4;
  node2->next = node3;

  node3->data = 3;
  node3->next = node4;

  node4->data = 2;
  node4->next = node5;

  node5->data = 1;
  node5->next = NULL;

  printf ("Before deletion: ");
  display (head);

  delete_Alt (head);

  printf ("After deletion: ");
  display (head);
  return 0;
}

Output

Before deletion: 5 4 3 2 1 
After deletion: 5 3 1