Delete Node at Beginning in DSA C# - Build from Scratch

Delete Node at Beginning in a Linked List

📖 Scenario: You are managing a simple list of tasks using a linked list. Each task is a node in the list. Sometimes, you need to remove the first task when it is completed.

🎯 Goal: Build a linked list with three tasks, then remove the first task from the list, and finally print the remaining tasks.

📋 What You'll Learn

Create a linked list with exactly three nodes containing the strings: "Task1", "Task2", "Task3"

Create a variable called head that points to the first node

Write code to delete the first node by updating head

Print the linked list nodes from head to the end, separated by ' -> ', ending with 'null'

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like task scheduling, undo functionality in editors, and managing playlists.

💼 Career

Understanding linked list operations like deletion is fundamental for software development roles involving data structures and memory management.

1

Create the linked list with three nodes

Create a class called Node with a string data and a Node reference called next. Then create three nodes with data "Task1", "Task2", and "Task3". Link them so that node1.next is node2 and node2.next is node3. Finally, create a variable called head that points to node1.

DSA C#

// Define Node class and create three linked nodes
// Your code here

Hint

Define a Node class with a constructor. Then create three nodes and link them by setting the next property.

2

Prepare to delete the first node

Create a variable called temp and set it to head. This will hold the node to be deleted.

DSA C#

public class Node
{
    public string data;
    public Node next;
    public Node(string data)
    {
        this.data = data;
        this.next = null;
    }
}

Node node1 = new Node("Task1");
Node node2 = new Node("Task2");
Node node3 = new Node("Task3");
node1.next = node2;
node2.next = node3;
Node head = node1;

// Create a variable temp and set it to head
// Your code here

Hint

Use Node temp = head; to keep track of the node you want to delete.

3

Delete the first node by updating head

Set head to head.next to remove the first node from the list.

DSA C#

public class Node
{
    public string data;
    public Node next;
    public Node(string data)
    {
        this.data = data;
        this.next = null;
    }
}

Node node1 = new Node("Task1");
Node node2 = new Node("Task2");
Node node3 = new Node("Task3");
node1.next = node2;
node2.next = node3;
Node head = node1;

Node temp = head;

// Update head to remove the first node
// Your code here

Hint

Assign head = head.next; to remove the first node from the list.

4

Print the linked list after deletion

Use a Node variable called current starting at head. Use a while loop to print each node's data followed by " -> ". After the loop, print "null".

DSA C#

public class Node
{
    public string data;
    public Node next;
    public Node(string data)
    {
        this.data = data;
        this.next = null;
    }
}

Node node1 = new Node("Task1");
Node node2 = new Node("Task2");
Node node3 = new Node("Task3");
node1.next = node2;
node2.next = node3;
Node head = node1;

Node temp = head;
head = head.next;

// Print the linked list starting from head
// Your code here

Hint

Use a while loop to go through each node starting at head. Print the data and " -> ". After the loop, print "null".

Practice

1. What happens when you delete the first node in a singly linked list?

easy

A. The last node is deleted instead of the first

B. Nothing happens because the first node cannot be deleted

C. The entire list is deleted

D. The head pointer moves to the second node, removing the first node from the list

Solution

1. Step 1: Understand the head pointer role

   The head pointer points to the first node in the list.

2. Step 2: Update head to next node

   Deleting the first node means moving the head pointer to the second node, effectively removing the first node.

3. Final Answer:

   The head pointer moves to the second node, removing the first node from the list -> Option D

4. Quick Check:

   Delete first node = move head to next [OK]

Hint: Delete first node by moving head to next node [OK]

Common Mistakes:

• Trying to delete the last node instead
• Not updating the head pointer
• Deleting the entire list accidentally

2. Which of the following is the correct syntax to delete the first node in a singly linked list in C#?

easy

A. head = null.next;

B. head.next = head;

C. head = head.next;

D. head.next = null;

Solution

1. Step 1: Identify the head pointer update

   To delete the first node, assign head to head.next to move the head pointer forward.

2. Step 2: Check syntax correctness

   head = head.next; correctly updates the head pointer to the second node.

3. Final Answer:

   head = head.next; -> Option C

4. Quick Check:

   Correct syntax to delete first node = head = head.next; [OK]

Hint: Assign head to head.next to delete first node [OK]

Common Mistakes:

• Reversing the assignment order
• Using null.next which is invalid
• Setting head.next to null instead of head

3. Consider the following code snippet for deleting the first node in a linked list:

Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head = head.next;

What is the state of the list after deletion?

medium

A. 2 -> 3 -> null

B. null

C. 3 -> null

D. 1 -> 2 -> 3 -> null

Solution

1. Step 1: Initial list creation

   The list starts as 1 -> 2 -> 3 -> null with head at node 1.

2. Step 2: Delete first node by moving head

   head = head.next moves head to node 2, so the list becomes 2 -> 3 -> null.

3. Final Answer:

   2 -> 3 -> null -> Option A

4. Quick Check:

   After deletion, list starts at second node [OK]

Hint: After deletion, list starts from second node [OK]

Common Mistakes:

• Forgetting to update head pointer
• Assuming first node still exists
• Confusing node values after deletion

4. The following code attempts to delete the first node of a linked list but causes a runtime error:

Node head = null;
head = head.next;

What is the error and how to fix it?

medium

A. SyntaxError; fix by adding semicolon

B. NullReferenceException; fix by checking if head is not null before deleting

C. IndexOutOfRangeException; fix by resizing list

D. No error; code works fine

Solution

1. Step 1: Identify cause of error

   head is null, so accessing head.next causes NullReferenceException.

2. Step 2: Fix by adding null check

   Check if head != null before assigning head = head.next to avoid error.

3. Final Answer:

   NullReferenceException; fix by checking if head is not null before deleting -> Option B

4. Quick Check:

   Null head causes error; check before delete [OK]

Hint: Always check head != null before deleting first node [OK]

Common Mistakes:

• Ignoring null checks causing runtime errors
• Assuming list always has nodes
• Confusing syntax errors with runtime errors

5. You have a singly linked list with nodes: 5 -> 10 -> 15 -> 20 -> null. You delete the first node twice using head = head.next; two times. What is the resulting list?

hard

A. 15 -> 20 -> null

B. 10 -> 15 -> 20 -> null

C. 20 -> null

D. null

Solution

1. Step 1: Initial list and first deletion

   List starts as 5 -> 10 -> 15 -> 20 -> null. After first deletion, head moves to 10, list is 10 -> 15 -> 20 -> null.

2. Step 2: Second deletion moves head again

   After second deletion, head moves to 15, list is 15 -> 20 -> null.

3. Final Answer:

   15 -> 20 -> null -> Option A

4. Quick Check:

   Two deletions remove first two nodes [OK]

Hint: Each deletion moves head to next node [OK]

Common Mistakes:

• Deleting only once instead of twice
• Confusing node values after multiple deletions
• Assuming list becomes empty too early