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C++ Program To Delete Alternate Nodes Of A Linked List

  • Last Updated : 16 Dec, 2021

Given a Singly Linked List, starting from the second node delete all alternate nodes of it. For example, if the given linked list is 1->2->3->4->5 then your function should convert it to 1->3->5, and if the given linked list is 1->2->3->4 then convert it to 1->3.

Method 1 (Iterative): 
Keep track of previous of the node to be deleted. First, change the next link of the previous node and iteratively move to the next node.


// C++ program to remove alternate 
// nodes of a linked list 
#include <bits/stdc++.h>
using namespace std;
// A linked list node 
class Node 
    int data; 
    Node *next; 
/* Deletes alternate nodes 
   of a list starting with head */
void deleteAlt(Node *head) 
    if (head == NULL) 
    /* Initialize prev and node 
       to be deleted */
    Node *prev = head; 
    Node *node = head->next; 
    while (prev != NULL && 
           node != NULL) 
        /* Change next link of previous 
           node */
        prev->next = node->next; 
        // Update prev and node 
        prev = prev->next; 
        if (prev != NULL) 
            node = prev->next; 
   fun1() and fun2() */
/* Given a reference (pointer to pointer) 
   to the head of a list and an int, push 
   a new node on the front of the list. */
void push(Node** head_ref, 
          int new_data) 
    /* Allocate node */
    Node* new_node = new Node();
    /* Put in the data */
    new_node->data = new_data; 
    /* Link the old list off the 
       new node */
    new_node->next = (*head_ref); 
    /* Move the head to point to the 
       new node */
    (*head_ref) = new_node; 
/* Function to print nodes in a 
   given linked list */
void printList(Node *node) 
    while (node != NULL) 
        cout << node->data << " "
        node = node->next; 
// Driver code 
int main() 
    // Start with the empty list 
    Node* head = NULL; 
    /* Using push() to construct list 
       1->2->3->4->5 */
    push(&head, 5); 
    push(&head, 4); 
    push(&head, 3); 
    push(&head, 2); 
    push(&head, 1); 
    cout << "List before calling deleteAlt() "
    cout << "List after calling deleteAlt() "
    return 0; 
// This code is contributed by rathbhupendra


List before calling deleteAlt() 
1 2 3 4 5 
List after calling deleteAlt() 
1 3 5 

Time Complexity: O(n) where n is the number of nodes in the given Linked List.

Method 2 (Recursive): 
Recursive code uses the same approach as method 1. The recursive code is simple and short but causes O(n) recursive function calls for a linked list of size n.


/* Deletes alternate nodes of a list 
   starting with head */
void deleteAlt(Node *head) 
    if (head == NULL) 
    Node *node = head->next; 
    if (node == NULL) 
    // Change the next link of head 
    head->next = node->next; 
    // Free memory allocated for node 
    /* Recursively call for the new 
       next of head */
// This code is contributed by rathbhupendra

Time Complexity: O(n)

Please refer complete article on Delete alternate nodes of a Linked List for more details!

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