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

Insert at End of Circular Linked List

📖 Scenario: You are managing a circular queue of tasks where the last task points back to the first task, forming a circle. 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 circular list to verify the insertion.

📋 What You'll Learn

Create a Node class with data and next attributes

Create a circular linked list with three nodes containing data 10, 20, and 30

Create a function insert_at_end(head, data) to insert a new node at the end

Print the circular linked list starting from head, showing all nodes once

💡 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 system programming, embedded systems, and designing efficient data structures.

1

Create Node class and initial circular linked list

Create a class called Node with attributes data and next. Then create three nodes with data 10, 20, and 30. Link them in a circular way so that the last node points back to the first node. Assign the first node to a variable called head.

DSA Python

# Define Node class and create circular linked list with nodes 10, 20, 30
# Your code here

Hint

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

2

Create a function to insert at end

Create a function called insert_at_end(head, data) that inserts a new node with the given data at the end of the circular linked list. The function should return the head of the updated list.

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)

node1.next = node2
node2.next = node3
node3.next = node1

head = node1

# Define insert_at_end function below
# Your code here

Hint

Traverse the list until you reach the last node (which points back to head), then insert the new node after it.

3

Insert a new node with data 40 at the end

Call the function insert_at_end(head, 40) and assign the result back to head to insert a new node with data 40 at the end of the circular linked list.

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)

node1.next = node2
node2.next = node3
node3.next = node1

head = node1

def insert_at_end(head, data):
    new_node = Node(data)
    if head is None:
        new_node.next = new_node
        return new_node
    current = head
    while current.next != head:
        current = current.next
    current.next = new_node
    new_node.next = head
    return head

# Insert new node with data 40 at the end
# Your code here

Hint

Assign the result of insert_at_end(head, 40) back to head.

4

Print the circular linked list

Write code to print the data of each node in the circular linked list starting from head. Print all nodes exactly once, separated by ->, and end with null.

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)

node1.next = node2
node2.next = node3
node3.next = node1

head = node1

def insert_at_end(head, data):
    new_node = Node(data)
    if head is None:
        new_node.next = new_node
        return new_node
    current = head
    while current.next != head:
        current = current.next
    current.next = new_node
    new_node.next = head
    return head

head = insert_at_end(head, 40)

# Print the circular linked list
# Your code here

Hint

Use a loop to traverse nodes until you reach the head again, collecting data to print.

Practice

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

easy

A. The new node is ignored because the list is circular.

B. The new node becomes the head node and points to itself only.

C. The new node is inserted at the beginning and breaks the circular link.

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

Solution

1. Step 1: Understand the 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 the end

   The new node is linked after the current last node, and its next pointer is set 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 D

4. Quick Check:

   Insert at end keeps circle intact = D [OK]

Hint: New end node always points to head to keep circle [OK]

Common Mistakes:

• Forgetting to link new node back to head
• Breaking the circular link by pointing to None
• Inserting at beginning instead of end

2. Which of the following is the correct syntax to create a new node and insert it at the end of a circular linked list in Python?

easy

A. new_node = Node(data); new_node.next = None; last.next = new_node

B. new_node = Node(data); new_node.next = head; last.next = new_node

C. new_node = Node(data); head = new_node; new_node.next = head

D. new_node = Node(data); last.next = None; new_node.next = last

Solution

1. Step 1: Create new node and link to head

   The new node's next pointer must point to the head to maintain circularity.

2. Step 2: Link last node to new node

   The current last node's next pointer is updated to point to the new node, completing the insertion.

3. Final Answer:

   new_node = Node(data); new_node.next = head; last.next = new_node -> Option B

4. Quick Check:

   New node points to head, last points to new node = A [OK]

Hint: New node next = head; last next = new node [OK]

Common Mistakes:

• Setting new node next to None breaks circularity
• Not updating last node's next pointer
• Assigning head to new node incorrectly

3. What will be the printed output of the circular linked list after inserting 4 at the end, given the initial list is 1 -> 2 -> 3 -> (back to 1)?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def insert_end(head, data):
    new_node = Node(data)
    if not head:
        new_node.next = new_node
        return new_node
    temp = head
    while temp.next != head:
        temp = temp.next
    temp.next = new_node
    new_node.next = head
    return head

head = Node(1)
second = Node(2)
third = Node(3)
head.next = second
second.next = third
third.next = head

head = insert_end(head, 4)

# Print circular list
current = head
output = []
while True:
    output.append(current.data)
    current = current.next
    if current == head:
        break
print(output)

medium

A. [1, 2, 3]

B. [4, 1, 2, 3]

C. [1, 2, 3, 4]

D. [1, 2, 3, 4, 1]

Solution

1. Step 1: Insert 4 at the end of the circular list

   The function finds the last node (3), links it to new node (4), and new node points back to head (1).

2. Step 2: Print nodes starting from head

   Traverse from head: 1 -> 2 -> 3 -> 4, then back to 1 stops loop. Output list is [1, 2, 3, 4].

3. Final Answer:

   [1, 2, 3, 4] -> Option C

4. Quick Check:

   List after insert = [1, 2, 3, 4] [OK]

Hint: Traverse until back to head to print full circular list [OK]

Common Mistakes:

• Printing extra repeated head node
• Not updating last node's next pointer
• Stopping traversal too early

4. Identify the error in this code snippet for inserting at the end of a circular linked list:

def insert_end(head, data):
    new_node = Node(data)
    if not head:
        new_node.next = None
        return new_node
    temp = head
    while temp.next != head:
        temp = temp.next
    temp.next = new_node
    new_node.next = head
    return head

medium

A. new_node.next should point to new_node when list is empty

B. temp.next should be set to None instead of new_node

C. The while loop condition should be temp.next == head

D. The function should return new_node instead of head

Solution

1. Step 1: Check empty list case

   When list is empty, new_node.next must point to itself to form a circular list, not None.

2. Step 2: Verify rest of code

   Other parts correctly insert new_node at end and maintain circularity.

3. Final Answer:

   new_node.next should point to new_node when list is empty -> Option A

4. Quick Check:

   Empty list node points to itself = B [OK]

Hint: Empty circular list node points to itself, not None [OK]

Common Mistakes:

• Setting new_node.next to None breaks circularity
• Wrong while loop condition
• Returning wrong node after insertion

5. Given a circular linked list with nodes 10 -> 20 -> 30 -> (back to 10), write the output after inserting 5 at the end and then inserting 40 at the end. What is the final list printed?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def insert_end(head, data):
    new_node = Node(data)
    if not head:
        new_node.next = new_node
        return new_node
    temp = head
    while temp.next != head:
        temp = temp.next
    temp.next = new_node
    new_node.next = head
    return head

head = Node(10)
second = Node(20)
third = Node(30)
head.next = second
second.next = third
third.next = head

head = insert_end(head, 5)
head = insert_end(head, 40)

current = head
output = []
while True:
    output.append(current.data)
    current = current.next
    if current == head:
        break
print(output)

hard

A. [10, 20, 30, 5, 40]

B. [5, 10, 20, 30, 40]

C. [10, 20, 30, 40, 5]

D. [40, 10, 20, 30, 5]

Solution

1. Step 1: Insert 5 at the end

   After first insertion, list becomes 10 -> 20 -> 30 -> 5 -> (back to 10).

2. Step 2: Insert 40 at the end

   After second insertion, list becomes 10 -> 20 -> 30 -> 5 -> 40 -> (back to 10).

3. Step 3: Print the list

   Traverse from head: 10, 20, 30, 5, 40, then back to 10 stops loop. Output is [10, 20, 30, 5, 40].

4. Final Answer:

   [10, 20, 30, 5, 40] -> Option A

5. Quick Check:

   Two inserts at end add nodes in order = A [OK]

Hint: Insertions add nodes at end, order preserved [OK]

Common Mistakes:

• Inserting new nodes at beginning by mistake
• Not updating last node's next pointer correctly
• Printing extra repeated nodes