Delete Node at End in DSA C#

Mental Model

To remove the last item from a chain, find the second last item and make it the new end by cutting off the last one.

Analogy: Imagine a line of people holding hands. To remove the last person, the second last person lets go of their hand, so the last person is no longer connected.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, delete node at end

Goal: Remove the last node (3) so the list ends at node 2

Step 1: start at head node (1)

head -> [curr->1] -> 2 -> 3 -> null

Why: We begin from the start to find the node before the last

Step 2: move curr to next node (2)

head -> 1 -> [curr->2] -> 3 -> null

Why: Keep moving forward to reach the second last node

Step 3: check next node's next: curr.next.next is null, so curr is second last

head -> 1 -> [curr->2] -> 3 -> null

Why: We found the node before the last node

Step 4: set curr.next to null, removing last node

head -> 1 -> 2 -> null

Why: Cut off the last node by making second last node the new end

Result:

head -> 1 -> 2 -> null

Annotated Code

DSA C#

using System;

public class Node
{
    public int data;
    public Node next;

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

public class LinkedList
{
    public Node head;

    public void InsertAtEnd(int data)
    {
        Node newNode = new Node(data);
        if (head == null)
        {
            head = newNode;
            return;
        }
        Node curr = head;
        while (curr.next != null)
        {
            curr = curr.next;
        }
        curr.next = newNode;
    }

    public void DeleteAtEnd()
    {
        if (head == null) // empty list
            return;

        if (head.next == null) // only one node
        {
            head = null;
            return;
        }

        Node curr = head;
        while (curr.next.next != null) // find second last node
        {
            curr = curr.next;
        }
        curr.next = null; // remove last node
    }

    public void PrintList()
    {
        Node curr = head;
        while (curr != null)
        {
            Console.Write(curr.data);
            if (curr.next != null)
                Console.Write(" -> ");
            curr = curr.next;
        }
        Console.WriteLine(" -> null");
    }
}

public class Program
{
    public static void Main()
    {
        LinkedList list = new LinkedList();
        list.InsertAtEnd(1);
        list.InsertAtEnd(2);
        list.InsertAtEnd(3);

        list.DeleteAtEnd();
        list.PrintList();
    }
}

if (head == null) return;

handle empty list by doing nothing

if (head.next == null) { head = null; return; }

handle single node list by removing head

while (curr.next.next != null) { curr = curr.next; }

advance curr to second last node

curr.next = null;

remove last node by cutting link

Output

1 -> 2 -> null

Complexity Analysis

Time: O(n) because we must traverse the list to find the second last node

Space: O(1) because we only use a few pointers regardless of list size

vs Alternative: Compared to removing from the front which is O(1), deleting at end requires traversal making it slower

Edge Cases

empty list

no change, safely returns without error

DSA C#

if (head == null) return;

single node list

head becomes null, list becomes empty

DSA C#

if (head.next == null) { head = null; return; }

When to Use This Pattern

When asked to remove the last item from a linked list, look for the pattern of finding the second last node to update its next pointer to null.

Common Mistakes

Mistake: Trying to delete the last node by traversing until curr.next is null and then setting curr to null
Fix: Instead, find the second last node and set its next pointer to null to remove the last node

Summary

Deletes the last node of a singly linked list by updating the second last node's next pointer to null.

Use when you need to remove the last element from a linked list.

The key insight is to stop at the second last node so you can cut off the last node cleanly.

Practice

1. What does the operation Delete Node at End do in a singly linked list?

easy

A. Adds a new node at the end of the list

B. Removes the first node from the list

C. Removes the last node from the list

D. Adds a new node at the beginning of the list

Solution

1. Step 1: Understand the operation meaning

   Deleting a node at the end means removing the last node in the linked list.

2. Step 2: Compare options with operation

   Only Removes the last node from the list matches removing the last node; others add or remove nodes elsewhere.

3. Final Answer:

   Removes the last node from the list -> Option C

4. Quick Check:

   Delete Node at End = Remove last node [OK]

Hint: Deleting at end means removing the last node [OK]

Common Mistakes:

• Confusing delete at end with delete at beginning
• Thinking it adds a node instead of removing
• Assuming it deletes a middle node

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

easy

A. Create a new node and assign it to last node

B. Set head to null directly

C. Set last node's next to head

D. Traverse to second last node, set its next to null

Solution

1. Step 1: Identify correct deletion method

   To delete last node, we must reach second last node and set its next pointer to null.

2. Step 2: Check options for this logic

   Traverse to second last node, set its next to null matches this logic; others do unrelated actions.

3. Final Answer:

   Traverse to second last node, set its next to null -> Option D

4. Quick Check:

   Delete last node = second last node next = null [OK]

Hint: Set second last node's next to null to delete last [OK]

Common Mistakes:

• Setting head to null deletes whole list
• Trying to set last node's next to head causes cycle
• Creating new node does not delete existing node

3. Given the linked list: 10 -> 20 -> 30 -> 40 -> null, what will be the list after deleting the node at the end?

medium

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

B. 10 -> 20 -> 30 -> null

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

D. 10 -> 20 -> null

Solution

1. Step 1: Identify last node in list

   The last node is 40 in the list 10 -> 20 -> 30 -> 40 -> null.

2. Step 2: Remove last node and update list

   Deleting last node removes 40, so list becomes 10 -> 20 -> 30 -> null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option B

4. Quick Check:

   Last node 40 removed = 10->20->30->null [OK]

Hint: Remove last value to get new list [OK]

Common Mistakes:

• Not removing last node, list unchanged
• Removing first node instead
• Removing more than one node accidentally

4. Consider this code snippet to delete the last node of a singly linked list:

Node current = head;
while(current.next.next != null) {
    current = current.next;
}
current.next = null;

What error can occur when this code runs on a list with only one node?

medium

A. NullReferenceException because current.next.next is invalid

B. No error, code works fine

C. Infinite loop because condition never false

D. Deletes the head node incorrectly

Solution

1. Step 1: Analyze condition for single node list

   If list has one node, current.next is null, so current.next.next causes error.

2. Step 2: Identify exception type

   Accessing next of null causes NullReferenceException in C#.

3. Final Answer:

   NullReferenceException because current.next.next is invalid -> Option A

4. Quick Check:

   Single node list causes null.next access error [OK]

Hint: Check for null before accessing next.next [OK]

Common Mistakes:

• Ignoring null check for single node
• Assuming loop runs safely always
• Thinking it deletes head node

5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> 35 -> 45 -> null. Write a method to delete the last node and return the updated list. Which of the following code snippets correctly implements this?

hard

A. if (head == null) return null;
if (head.next == null) return null;
Node current = head;
while (current.next.next != null) {
current = current.next;
}
current.next = null;
return head;

B. if (head == null) return null;
Node current = head;
while (current.next != null) {
current = current.next;
}
current = null;
return head;

C. if (head == null) return null;
Node current = head;
while (current.next.next != null) {
current = current.next;
}
current.next = current;
return head;

D. if (head == null) return null;
head = head.next;
return head;

Solution

1. Step 1: Handle empty and single node cases

   If list is empty or has one node, deleting last node returns null (empty list).

2. Step 2: Traverse to second last node and delete last

   Loop until current.next.next is null, then set current.next to null to remove last node.

3. Step 3: Return updated head

   Return head pointer to updated list.

4. Final Answer:

   Code snippet D correctly deletes last node and returns updated list -> Option A

5. Quick Check:

   Check null, traverse to second last, set next null [OK]

Hint: Check list size, then remove last by second last next = null [OK]

Common Mistakes:

• Setting current to null does not remove last node
• Assigning current.next to current creates cycle
• Deleting head instead of last node