Delete a Node from Circular Linked List in DSA C

Mental Model

A circular linked list loops back to the start, so deleting a node means adjusting pointers to skip that node while keeping the loop intact.

Analogy: Imagine people standing in a circle holding hands. Removing one person means the two neighbors hold hands directly, keeping the circle unbroken.

head -> 1 -> 2 -> 3 -> 4 -> (back to head 1)
          ↑                        ↓
          ←-----------------------

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> 4 (circular), delete node with value 3

Goal: Remove node 3 and keep the list circular without breaking the loop

Step 1: start at head node 1, look for node with value 3

head -> [1] -> 2 -> 3 -> 4 -> (back to head 1)

Why: We need to find the node to delete by traversing the list

Step 2: move to node 2, check if next node is 3

head -> 1 -> [2] -> 3 -> 4 -> (back to head 1)

Why: Check next node to find the node to delete

Step 3: found node 3 as next of 2, update node 2's next to node 4

head -> 1 -> 2 -> [4] -> (back to head 1)

Why: By skipping node 3, we remove it from the circle

Step 4: node 3 is removed, list remains circular: 1 -> 2 -> 4 -> (back to 1)

head -> 1 -> 2 -> 4 -> (back to head 1)

Why: The circle is intact without node 3

Result:

head -> 1 -> 2 -> 4 -> (back to head 1)

Annotated Code

DSA C

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

// Node structure for circular linked list
typedef struct Node {
    int data;
    struct Node* next;
} Node;

// Function to create a new node
Node* createNode(int data) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = data;
    newNode->next = newNode; // points to itself initially
    return newNode;
}

// Function to insert node at end of circular linked list
void insertEnd(Node** head, int data) {
    Node* newNode = createNode(data);
    if (*head == NULL) {
        *head = newNode;
        return;
    }
    Node* temp = *head;
    while (temp->next != *head) {
        temp = temp->next; // advance to last node
    }
    temp->next = newNode; // last node points to new node
    newNode->next = *head; // new node points to head
}

// Function to delete node with given value from circular linked list
void deleteNode(Node** head, int key) {
    if (*head == NULL) return; // empty list

    Node *curr = *head, *prev = NULL;

    // If head node itself holds the key and list has only one node
    if (curr->data == key && curr->next == *head) {
        free(curr);
        *head = NULL;
        return;
    }

    // If head node holds the key and list has more than one node
    if (curr->data == key) {
        // Find last node to update its next pointer
        Node* last = *head;
        while (last->next != *head) {
            last = last->next; // advance to last node
        }
        last->next = curr->next; // last node points to second node
        *head = curr->next; // head updated to second node
        free(curr); // delete old head
        return;
    }

    // Search for the node to delete
    prev = curr;
    curr = curr->next;
    while (curr != *head && curr->data != key) {
        prev = curr;
        curr = curr->next; // advance curr and prev
    }

    // If node with key found
    if (curr != *head && curr->data == key) {
        prev->next = curr->next; // bypass curr
        free(curr); // delete node
    }
}

// Function to print circular linked list
void printList(Node* head) {
    if (head == NULL) {
        printf("List is empty\n");
        return;
    }
    Node* temp = head;
    do {
        printf("%d -> ", temp->data);
        temp = temp->next; // advance to next node
    } while (temp != head);
    printf("(back to head %d)\n", head->data);
}

int main() {
    Node* head = NULL;
    insertEnd(&head, 1);
    insertEnd(&head, 2);
    insertEnd(&head, 3);
    insertEnd(&head, 4);

    printf("Original list:\n");
    printList(head);

    deleteNode(&head, 3);

    printf("After deleting node with value 3:\n");
    printList(head);

    return 0;
}

while (temp->next != *head) { temp = temp->next; }

advance temp to last node to insert new node at end

if (curr->data == key && curr->next == *head) { free(curr); *head = NULL; return; }

handle deletion when only one node exists and it matches key

if (curr->data == key) { find last node; update last->next; update head; free old head; return; }

handle deletion when head node matches key and list has multiple nodes

while (curr != *head && curr->data != key) { prev = curr; curr = curr->next; }

search for node with key by traversing list

if (curr != *head && curr->data == key) { prev->next = curr->next; free(curr); }

remove found node by bypassing it and freeing memory

Output

Original list:

1 -> 2 -> 3 -> 4 -> (back to head 1)

After deleting node with value 3:

1 -> 2 -> 4 -> (back to head 1)

Complexity Analysis

Time: O(n) because in worst case we traverse all nodes once to find the node to delete

Space: O(1) because we only use a few pointers and no extra data structures

vs Alternative: Compared to singly linked list deletion which may require special handling for tail, circular list deletion keeps traversal simple but requires careful pointer updates to maintain the loop

Edge Cases

empty list

function returns immediately without changes

DSA C

if (*head == NULL) return;

list with one node which is the node to delete

node is freed and head set to NULL

DSA C

if (curr->data == key && curr->next == *head) { free(curr); *head = NULL; return; }

deleting the head node in a list with multiple nodes

head pointer updated and last node's next pointer updated to new head

DSA C

if (curr->data == key) { find last node; update last->next; update head; free old head; return; }

node with key not found

list remains unchanged

DSA C

while (curr != *head && curr->data != key) { ... }

When to Use This Pattern

When you see a problem asking to delete a node from a circular linked list, reach for pointer adjustment pattern because you must keep the loop intact while removing the node.

Common Mistakes

Mistake: Not updating the last node's next pointer when deleting the head node
Fix: Find the last node and update its next pointer to the new head before deleting the old head

Mistake: Not handling the case when the list has only one node
Fix: Check if the node to delete is the only node and set head to NULL after deletion

Mistake: Stopping traversal too early or infinite loop by not checking circular condition
Fix: Use a do-while or carefully check if current node equals head again to stop traversal

Summary

Deletes a node with a given value from a circular linked list while keeping the list circular.

Use when you need to remove elements from a circular linked list without breaking the loop.

The key is to adjust pointers carefully so the circle remains unbroken after deletion.