Insert at End of Circular Linked List in DSA C - Execution Trace

Concept Flow - Insert at End of Circular Linked List

Create new node

↓

Is list empty?

Yes→new_node.next = new_node

↓

head = new_node

↓

Start at head

↓

Traverse until current.next == head

↓

Set current.next = new_node

↓

Set new_node.next = head

↓

Done

Create a new node, if list empty point it to itself. Otherwise, traverse to last node, link it to new node, and new node back to head.

Execution Sample

DSA C

struct Node {
  int data;
  struct Node* next;
};

void insertEnd(struct Node** head, int val) {
  struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));
  new_node->data = val;
  new_node->next = NULL;

  if (*head == NULL) {
    new_node->next = new_node;
    *head = new_node;
    return;
  }

  struct Node* current = *head;
  while (current->next != *head) {
    current = current->next;
  }
  current->next = new_node;
  new_node->next = *head;
}

This code inserts a new node with value val at the end of a circular linked list.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Create new node with data=10	empty	new_node created, next=∅	No nodes
2	Check if list empty (head == NULL)	empty	Yes, new_node.next = new_node	┌────────┐ │ data:10│ │ next:─┐│ └───────┘│ └─→ (points to itself)
3	Set head = new_node	1 node	head → new_node	┌────────┐ │ data:10│ │ next:─┐│ └───────┘│ └─→ (points to itself) head
4	Insert 20 at end: create new node	1 node	new_node created with data=20, next=∅	┌────────┐ │ data:10│ │ next:─┐│ └───────┘│ └─→ (points to itself) head new_node: ┌────────┐ │ data:20│ │ next: ∅│ └───────┘
5	Traverse to last node (node with next=head)	1 node	current at node 10	┌────────┐ │ data:10│ │ next:─┐│ └───────┘│ └─→ head head
6	Set current.next = new_node	2 nodes	node 10.next → node 20	┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ next:─┐│──→ │ next: ∅│ └───────┘│ └───────┘ │ head └─────────────→ ?
7	Set new_node.next = head	2 nodes	node 20.next → head (node 10)	┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ next:─┐│──→ │ next:─┐│ └───────┘│ └───────┘│ │ └─→ head head └───────────────→
8	Insert 30 at end: create new node	2 nodes	new_node created with data=30, next=∅	List: ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ next:─┐│──→ │ next:─┐│ └───────┘│ └───────┘│ │ └─→ head head └───────────────→ new_node: ┌────────┐ │ data:30│ │ next: ∅│ └───────┘
9	Traverse to last node (node with next=head)	2 nodes	current moves: 10 → 20	┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ next:─┐│──→ │ next:─┐│ └───────┘│ └───────┘│ │ └─→ head head └───────────────→ current at node 20
10	Set current.next = new_node	3 nodes	node 20.next → node 30	┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ next:─┐│──→ │ next:─┐│──→ │ next: ∅│ └───────┘│ └───────┘│ └───────┘ │ │ head └──────────────┘
11	Set new_node.next = head	3 nodes	node 30.next → head (node 10)	┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ next:─┐│──→ │ next:─┐│──→ │ next:─┐│ └───────┘│ └───────┘│ └───────┘│ │ │ └─→ head head └──────────────┘
12	End of insertions	3 nodes	No pointer changes	Final circular list: ┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ next:─┐│──→ │ next:─┐│──→ │ next:─┐│ └───────┘│ └───────┘│ └───────┘│ │ │ └─→ head head └──────────────┘

💡 Insertion ends after linking new node's next to head, maintaining circular structure.

Variable Tracker

Variable	Start	After Step 3	After Step 6	After Step 7	After Step 10	After Step 11	Final
head	NULL	Node(10)	Node(10)	Node(10)	Node(10)	Node(10)	Node(10)
new_node	NULL	Node(10)	Node(20)	Node(20)	Node(30)	Node(30)	NULL
current	NULL	NULL	Node(10)	Node(10)	Node(20)	Node(20)	NULL
list_size	0	1	1	2	2	3	3

Key Moments - 3 Insights

Why does the new node point back to head instead of NULL?

Because it's a circular linked list, the last node must point back to head to keep the circle. See execution_table rows 7 and 11 where new_node.next is set to head.

Why do we traverse until current.next == head, not current == NULL?

In circular lists, there is no NULL at the end. The last node points back to head. So we stop when current.next is head, as shown in steps 5 and 9.

What happens if the list is empty before insertion?

We create a new node that points to itself and set head to it, making a single-node circular list. See steps 2 and 3.

Visual Quiz - 3 Questions

Look at the execution_table at step 7, what does new_node.next point to?

A NULL

B head node (data=10)

C previous last node (data=10)

D new_node itself

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns at step 7.

At which step does the list first have two nodes?

A Step 3

B Step 6

C Step 7

D Step 10

💡 Hint

Look at 'Nodes in List' and 'Pointer Changes' columns to see when second node is linked.

If the list was empty, what pointer does head point to after first insertion?

A new_node itself

B next node (which is NULL)

C NULL

D previous node

💡 Hint

Refer to variable_tracker and execution_table steps 2 and 3.

Concept Snapshot

Insert at End of Circular Linked List:
- Create new node
- If list empty: new_node.next = new_node, head = new_node
- Else traverse to last node (current.next == head)
- Set current.next = new_node
- Set new_node.next = head
- Maintains circular link structure

Full Transcript

To insert at the end of a circular linked list, first create a new node with the given data. If the list is empty (head is NULL), point the new node's next to itself and set head to this new node, forming a single-node circle. If the list is not empty, start at head and traverse nodes until you find the last node whose next points back to head. Change this last node's next pointer to the new node. Then set the new node's next pointer back to head, preserving the circular structure. This process ensures the new node is added at the end and the list remains circular.

Practice

1. What happens when you insert a new node at the end of a circular linked list that currently has 3 nodes?

easy

A. The new node becomes the head node and points to the second node.

B. The new node becomes the last node and points back to the head node.

C. The new node is inserted in the middle of the list.

D. The list breaks and becomes a linear linked list.

Solution

1. Step 1: Understand circular linked list structure

   In a circular linked list, the last node points back to the head node, forming a circle.

2. Step 2: Insert new node at end

   The new node is linked after the current last node, and its next pointer points to the head node, maintaining the circle.

3. Final Answer:

   The new node becomes the last node and points back to the head node. -> Option B

4. Quick Check:

   New node points to head, preserving circular link [OK]

Hint: New end node always points to head in circular list [OK]

Common Mistakes:

• Forgetting to link new node back to head
• Inserting node at wrong position
• Breaking the circular link

2. Which of the following is the correct syntax to create a new node in C for a circular linked list?

easy

A. struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));

B. newNode = malloc(sizeof(Node));

C. struct Node newNode = malloc(sizeof(struct Node));

D. Node* newNode = new Node();

Solution

1. Step 1: Allocate memory for new node

   In C, use malloc with sizeof(struct Node) and cast to (struct Node*).

2. Step 2: Correct pointer declaration

   Declare newNode as a pointer to struct Node to hold the allocated memory address.

3. Final Answer:

   struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); -> Option A

4. Quick Check:

   Proper malloc and pointer cast in C [OK]

Hint: Use malloc with sizeof(struct Node) and cast to struct Node* [OK]

Common Mistakes:

• Missing (struct Node*) cast
• Declaring newNode as struct Node instead of pointer
• Using C++ style new operator in C

3. Given the following code snippet for inserting at the end of a circular linked list, what will be the printed list after insertion?

struct Node {
  int data;
  struct Node* next;
};

void printList(struct Node* head) {
  struct Node* temp = head;
  if (head != NULL) {
    do {
      printf("%d -> ", temp->data);
      temp = temp->next;
    } while (temp != head);
  }
  printf("NULL\n");
}

int main() {
  struct Node* head = (struct Node*)malloc(sizeof(struct Node));
  struct Node* second = (struct Node*)malloc(sizeof(struct Node));
  struct Node* third = (struct Node*)malloc(sizeof(struct Node));

  head->data = 1; head->next = second;
  second->data = 2; second->next = third;
  third->data = 3; third->next = head;

  // Insert new node with data 4 at end
  struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
  newNode->data = 4;
  newNode->next = head;
  third->next = newNode;

  printList(head);
  return 0;
}

medium

A. 1 -> 2 -> 3 -> 4 -> 1 -> ... (infinite loop)

B. 1 -> 2 -> 3 -> NULL

C. 3 -> 4 -> 1 -> 2 -> NULL

D. 1 -> 2 -> 3 -> 4 -> NULL

Solution

1. Step 1: Understand list before insertion

   Initial list is 1 -> 2 -> 3 -> back to 1 (circular).

2. Step 2: Insert new node at end

   New node with data 4 points to head; third node points to new node, extending list to 1 -> 2 -> 3 -> 4 -> back to 1.

3. Step 3: Print list

   printList prints nodes until it reaches head again, so output is 1 -> 2 -> 3 -> 4 -> NULL.

4. Final Answer:

   1 -> 2 -> 3 -> 4 -> NULL -> Option D

5. Quick Check:

   New node added at end, circular link maintained [OK]

Hint: New node points to head; last node points to new node [OK]

Common Mistakes:

• Printing infinite loop by not stopping at head
• Forgetting to update last node's next pointer
• Incorrect print format

4. Identify the error in the following code snippet that attempts to insert a new node at the end of a circular linked list:

void insertEnd(struct Node** head, int data) {
  struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
  newNode->data = data;
  newNode->next = *head;

  struct Node* temp = *head;
  while (temp->next != NULL) {
    temp = temp->next;
  }
  temp->next = newNode;
}

medium

A. The function should not use double pointer for head.

B. newNode->next should be NULL instead of *head.

C. The while loop condition should check for temp->next != *head instead of NULL.

D. Memory allocation for newNode is missing.

Solution

1. Step 1: Analyze loop condition in circular list

   In a circular linked list, last node's next points to head, not NULL.

2. Step 2: Correct loop condition

   Loop should continue until temp->next equals head, not NULL, to find last node.

3. Final Answer:

   The while loop condition should check for temp->next != *head instead of NULL. -> Option C

4. Quick Check:

   Loop must detect circular end by comparing to head [OK]

Hint: In circular list, last node points to head, not NULL [OK]

Common Mistakes:

• Using NULL to detect end in circular list
• Setting newNode->next to NULL
• Not updating last node's next pointer

5. You have a circular linked list with only one node where head points to itself. You want to insert a new node at the end. Which sequence of steps correctly updates the list?

hard

A. Create new node, set newNode->next = head->next, set head->next = newNode.

B. Create new node, set newNode->next = head, set head->next = newNode, then update head to newNode.

C. Create new node, set newNode->next = NULL, set head->next = newNode.

D. Create new node, set newNode->next = newNode, set head = newNode.

Solution

1. Step 1: Understand single node circular list

   The single node has head->next pointing to itself (head).

2. Step 2: Insert at end

   Set newNode->next = head->next (which is head), then head->next = newNode. This makes head -> newNode -> head, maintaining circularity without changing head.

3. Final Answer:

   Create new node, set newNode->next = head->next, set head->next = newNode. -> Option A

4. Quick Check:

   head -> newNode -> head [OK]

Hint: For single node circular: newNode->next = head->next; head->next = newNode [OK]

Common Mistakes:

• Setting newNode->next to NULL breaking circularity
• Unnecessarily updating head to newNode
• Incorrectly linking new node to itself