Delete all the even nodes of a Circular Linked List

Given a circular singly linked list containing N nodes, the task is to delete all the even nodes from the list.


Input : 57->11->2->56->12->61 
Output : List after deletion : 57 -> 11 -> 61 

Input : 9->11->32->6->13->20
Output : List after deletion : 9 -> 11 -> 13 

The idea is to traverse the nodes of the circular singly linked list one by one and get the pointer of the nodes having even data. Delete those nodes by following the approach used in this post.

Below is the implementation of the above idea:





// CPP program to delete all even
// node from a Circular singly linked list
#include <bits/stdc++.h>
using namespace std;
// Structure for a node
struct Node {
    int data;
    struct Node* next;
// Function to insert a node at the begining
// of a Circular linked list
void push(struct Node** head_ref, int data)
    struct Node* ptr1 = (struct Node*)malloc(sizeof(struct Node));
    struct Node* temp = *head_ref;
    ptr1->data = data;
    ptr1->next = *head_ref;
    // If linked list is not NULL then
    // set the next of last node
    if (*head_ref != NULL) {
        while (temp->next != *head_ref)
            temp = temp->next;
        temp->next = ptr1;
        ptr1->next = ptr1; // For the first node
    *head_ref = ptr1;
// Delete the node if it is even
void deleteNode(Node* head_ref, Node* del)
    struct Node* temp = head_ref;
    // If node to be deleted is head node
    if (head_ref == del)
        head_ref = del->next;
    // travers list till not found
    // delete node
    while (temp->next != del) {
        temp = temp->next;
    // copy address of node
    temp->next = del->next;
    // Finally, free the memory occupied by del
// Function to delete all even nodes
// from the singly circular  linked list
void deleteEvenNodes(Node* head)
    struct Node* ptr = head;
    struct Node* next;
    // traverse list till the end
    // if the node is even then delete it
    do {
        // if node is even
        if (ptr->data % 2 == 0)
            deleteNode(head, ptr);
        // point to next node
        next = ptr->next;
        ptr = next;
    } while (ptr != head);
// Function to print nodes
void printList(struct Node* head)
    struct Node* temp = head;
    if (head != NULL) {
        do {
            printf("%d ", temp->data);
            temp = temp->next;
        } while (temp != head);
// Driver code
int main()
    // Initialize lists as empty
    struct Node* head = NULL;
    // Created linked list will be 57->11->2->56->12->61
    push(&head, 61);
    push(&head, 12);
    push(&head, 56);
    push(&head, 2);
    push(&head, 11);
    push(&head, 57);
    cout << "\nList after deletion : ";
    return 0;



List after deletion : 57 11 61

Time Complexity: O(N), where N is the total number of nodes.

My Personal Notes arrow_drop_up

Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

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