Insert at End of Circular Linked List in DSA C++ - Build from Scratch

Insert at End of Circular Linked List

📖 Scenario: You are managing a circular linked list that represents a round-robin queue of tasks. Each node holds a task number. You want to add a new task at the end of this circular list.

🎯 Goal: Build a circular linked list with initial tasks, then insert a new task at the end, and finally print the list to see the updated order.

📋 What You'll Learn

Create a circular linked list with exactly three nodes having values 10, 20, and 30.

Create a pointer variable called tail that points to the last node in the list.

Write a function called insertAtEnd that takes tail and an integer data to insert a new node at the end of the circular linked list.

Print the circular linked list starting from the node after tail to show all nodes in order.

💡 Why This Matters

🌍 Real World

Circular linked lists are used in real-time systems like task scheduling where the tasks repeat in a cycle.

💼 Career

Understanding circular linked lists helps in roles involving embedded systems, operating systems, and performance-critical applications.

1

Create initial circular linked list

Create a struct called Node with an int data and a Node* next. Then create three nodes with values 10, 20, and 30. Link them in a circular way so that the last node points back to the first node. Finally, create a pointer called tail that points to the last node (with value 30).

DSA C++

// Define Node struct
// Create three nodes with data 10, 20, 30
// Link nodes in circular fashion
// Set tail to last node

Hint

Remember to link the last node's next to the first node to make the list circular.

2

Add insertAtEnd function declaration

Declare a function called insertAtEnd that takes two parameters: a pointer to Node*& tail and an int data. This function will insert a new node with the given data at the end of the circular linked list. Add the function declaration above main.

DSA C++

#include 

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

// Declare insertAtEnd function here

int main() {
    Node* first = new Node{10, nullptr};
    Node* second = new Node{20, nullptr};
    Node* third = new Node{30, nullptr};

    first->next = second;
    second->next = third;
    third->next = first;

    Node* tail = third;

    return 0;
}

Hint

Use Node*& tail to allow the function to modify the tail pointer.

3

Implement insertAtEnd function

Implement the insertAtEnd function. If tail is nullptr, create a new node that points to itself and assign it to tail. Otherwise, create a new node, set its next to tail->next, then set tail->next to this new node, and finally update tail to the new node.

DSA C++

#include 

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

void insertAtEnd(Node*& tail, int data) {
    // Your code here
}

int main() {
    Node* first = new Node{10, nullptr};
    Node* second = new Node{20, nullptr};
    Node* third = new Node{30, nullptr};

    first->next = second;
    second->next = third;
    third->next = first;

    Node* tail = third;

    return 0;
}

Hint

Remember to handle the empty list case where tail is nullptr.

4

Insert new node and print list

Call insertAtEnd(tail, 40) to add a new node with value 40 at the end. Then print the circular linked list starting from tail->next and continue until you reach tail again, printing each node's data separated by spaces.

DSA C++

#include 

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

void insertAtEnd(Node*& tail, int data) {
    Node* newNode = new Node{data, nullptr};
    if (tail == nullptr) {
        newNode->next = newNode;
        tail = newNode;
    } else {
        newNode->next = tail->next;
        tail->next = newNode;
        tail = newNode;
    }
}

int main() {
    Node* first = new Node{10, nullptr};
    Node* second = new Node{20, nullptr};
    Node* third = new Node{30, nullptr};

    first->next = second;
    second->next = third;
    third->next = first;

    Node* tail = third;

    // Insert new node with data 40
    // Print the circular linked list

    return 0;
}

Hint

Use a do-while loop to print all nodes starting from tail->next until you come back to it.

Practice

1. What happens when you insert a new node at the end of an empty circular linked list?

easy

A. The new node points to itself, becoming the only node in the list.

B. The new node points to null, ending the list.

C. The new node is discarded because the list is empty.

D. The new node points to the head, but the list remains empty.

Solution

1. Step 1: Understand circular linked list empty state

   When the list is empty, there is no node, so the new node must point to itself to form a circle.

2. Step 2: Insert new node and update pointers

   The new node's next pointer is set to itself, making it the only node and the head of the list.

3. Final Answer:

   The new node points to itself, becoming the only node in the list. -> Option A

4. Quick Check:

   Empty list insertion creates a self-pointing node [OK]

Hint: New node points to itself if list is empty [OK]

Common Mistakes:

• Setting next pointer to null instead of itself
• Not updating the head pointer
• Assuming list remains empty after insertion

2. Which of the following is the correct syntax to insert a new node at the end of a circular linked list in C++?

easy

A. newNode->next = head; tail->next = newNode; tail = newNode;

B. newNode->next = nullptr; tail->next = newNode; tail = newNode;

C. newNode->next = tail; head->next = newNode; tail = newNode;

D. newNode->next = head; head->next = newNode; tail = newNode;

Solution

1. Step 1: Set new node's next to head

   In a circular linked list, the last node points to the head, so newNode->next = head.

2. Step 2: Update current tail's next and tail pointer

   Set tail->next to newNode to link the last node to new node, then update tail to newNode.

3. Final Answer:

   newNode->next = head; tail->next = newNode; tail = newNode; -> Option A

4. Quick Check:

   New node points to head, tail updated [OK]

Hint: Tail's next must point to new node, new node points to head [OK]

Common Mistakes:

• Setting newNode->next to nullptr instead of head
• Updating head->next instead of tail->next
• Not updating tail pointer after insertion

3. Given the circular linked list with nodes 10 -> 20 -> 30 -> (back to 10), what will be the list after inserting 40 at the end?

medium

A. 40 -> 10 -> 20 -> 30 -> (back to 40)

B. 10 -> 20 -> 30 -> (back to 40)

C. 10 -> 20 -> 30 -> 40 -> null

D. 10 -> 20 -> 30 -> 40 -> (back to 10)

Solution

1. Step 1: Identify current circular list structure

   The list is 10 -> 20 -> 30, and 30 points back to 10 forming a circle.

2. Step 2: Insert 40 at end and update pointers

   New node 40 points to head (10), old tail (30) points to 40, making list 10 -> 20 -> 30 -> 40 -> back to 10.

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> (back to 10) -> Option D

4. Quick Check:

   New node added at end, circle maintained [OK]

Hint: New node points to head, old tail points to new node [OK]

Common Mistakes:

• Breaking the circle by pointing last node to null
• Inserting new node at start instead of end
• Not updating tail pointer

4. What is wrong with this code snippet for inserting at the end of a circular linked list?

void insertEnd(Node*& tail, int data) {
  Node* newNode = new Node(data);
  newNode->next = tail->next;
  tail = newNode;
}

medium

A. It sets newNode->next to null instead of head.

B. It deletes the old tail node accidentally.

C. It does not update the old tail's next pointer to newNode.

D. It inserts the new node at the beginning instead of the end.

Solution

1. Step 1: Check pointer updates in code

   The code sets newNode->next to tail->next (head), but does not update tail->next to newNode.

2. Step 2: Identify missing link update

   Without updating old tail's next pointer, the list breaks the circle and newNode is not linked properly.

3. Final Answer:

   It does not update the old tail's next pointer to newNode. -> Option C

4. Quick Check:

   Old tail's next must point to new node [OK]

Hint: Always update old tail's next to new node [OK]

Common Mistakes:

• Forgetting to update old tail's next pointer
• Setting newNode->next incorrectly
• Not updating tail pointer after insertion

5. You have a circular linked list with nodes 5 -> 15 -> 25 -> (back to 5). You want to insert a new node with value 35 at the end. After insertion, you mistakenly forget to update the tail pointer to the new node. What will be the effect on the list?

hard

A. The list will have 5 -> 15 -> 25 -> (back to 5) without 35.

B. The list will have 5 -> 15 -> 25 -> 35 -> (back to 5), but tail still points to 25.

C. The new node 35 will not be linked and will be lost.

D. The list will break and become linear ending at 25.

Solution

1. Step 1: Insert new node and link old tail

   New node 35 is linked after 25, and points back to head (5), so list is 5 -> 15 -> 25 -> 35 -> back to 5.

2. Step 2: Tail pointer not updated

   Tail still points to 25, so operations relying on tail may behave incorrectly, but list structure is intact.

3. Final Answer:

   The list will have 5 -> 15 -> 25 -> 35 -> (back to 5), but tail still points to 25. -> Option B

4. Quick Check:

   Tail pointer must update, else points to old last node [OK]

Hint: Tail pointer must update to new node after insertion [OK]

Common Mistakes:

• Assuming list breaks without tail update
• Thinking new node is lost if tail not updated
• Confusing tail pointer with node links