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

Last Updated : 23 Jul, 2025
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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++ 
// C++ program to remove alternate 
// nodes of a linked list 
#include <bits/stdc++.h>
using namespace std;

// A linked list node 
class Node 
{ 
    public:
    int data; 
    Node *next; 
}; 

/* Deletes alternate nodes 
   of a list starting with head */
void deleteAlt(Node *head) 
{ 
    if (head == NULL) 
        return; 

    /* 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; 
    } 
} 

/* UTILITY FUNCTIONS TO TEST 
   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 of 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() "; 
    printList(head); 

    deleteAlt(head); 

    cout << "List after calling deleteAlt() "; 
    printList(head); 

    return 0; 
} 
// This code is contributed by rathbhupendra

Output: 

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.

Auxiliary Space: O(1) because it is using constant space

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.

C++ 
/* Deletes alternate nodes of a list 
   starting with head */
void deleteAlt(Node *head) 
{ 
    if (head == NULL) 
        return; 

    Node *node = head->next; 

    if (node == NULL) 
        return; 

    // Change the next link of head 
    head->next = node->next; 

    // Free memory allocated for node 
    free(node); 

    /* Recursively call for the new 
       next of head */
    deleteAlt(head->next); 
} 
// This code is contributed by rathbhupendra

Time Complexity: O(n)

Auxiliary space: O(n) for call stack because using recursion

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


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