Delete Node by Value in DSA C# - Execution Trace

Concept Flow - Delete Node by Value

Start at head

↓

Is head null?

Yes→End: List empty

No↓

Is head value == target?

Yes→Delete head

↓

Update head

↓

Traverse list

↓

Find node with target value?

No→End: Value not found

Yes↓

Delete node by linking previous to next

↓

End

Start from the head node, check if it matches the value to delete. If yes, remove head. Otherwise, traverse to find the node and remove it by adjusting links.

Execution Sample

DSA C#

class Node {
  public int val;
  public Node next;
  public Node(int v) { val = v; next = null; }
}

void DeleteNode(ref Node head, int target) {
  // delete node with value target
}

Deletes the first node with the given value from a singly linked list.

Execution Table

Step	Current Node Value	Previous Node Value	Condition	Action	List State
1	1	null	head.val == 3? No	Move to next node	1 -> 3 -> 5 -> 3 -> null
2	3	1	current.val == 3? Yes	Delete current node by linking previous.next to current.next	1 -> 5 -> 3 -> null
3	5	1	Stop traversal after deletion	End	1 -> 5 -> 3 -> null

💡 Node with value 3 found at step 2 and deleted; traversal stops.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
current	1 (head)	3	5	5
previous	null	1	1	1
head	1	1	1	1
list	1 -> 3 -> 5 -> 3 -> null	1 -> 3 -> 5 -> 3 -> null	1 -> 5 -> 3 -> null	1 -> 5 -> 3 -> null

Key Moments - 3 Insights

Why do we need to keep track of the previous node?

Because to delete a node in a singly linked list, we must change the previous node's next pointer to skip the current node. See execution_table step 2 where previous.next is updated.

What happens if the node to delete is the head?

If the head node matches the target value, we delete it by updating the head pointer to head.next. This is shown in the concept_flow before traversal.

Why do we stop traversal after deleting the node?

Because the problem deletes only the first occurrence of the value. After deletion, continuing traversal is unnecessary. See execution_table step 3.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'previous' after step 2?

A 1

B 3

C 5

D null

💡 Hint

Check the 'previous' column in execution_table row for step 2.

At which step does the node get deleted?

A Step 1

B Step 2

C Step 3

D No deletion occurs

💡 Hint

Look at the 'Action' column in execution_table to find when deletion happens.

If the target value was 1, what would happen at step 1?

A Delete head node and update head

B Move to next node

C Delete second node

D No deletion

💡 Hint

Refer to concept_flow where head node deletion is handled before traversal.

Concept Snapshot

Delete Node by Value in singly linked list:
- Start at head node
- If head matches value, update head to head.next
- Else traverse list keeping track of previous node
- When node with value found, link previous.next to current.next
- Stops after deleting first matching node
- Handles empty list and value not found gracefully

Full Transcript

To delete a node by value in a singly linked list, start at the head. If the head node contains the target value, delete it by moving the head pointer to the next node. Otherwise, traverse the list while keeping track of the previous node. When you find the node with the target value, delete it by linking the previous node's next pointer to the current node's next pointer. Stop traversal after deleting the first matching node. If the list is empty or the value is not found, no changes are made.

Practice

1. What does the operation Delete Node by Value do in a linked list?

easy

A. Adds a new node with the given value at the end of the list

B. Finds the node with the given value and returns it without removing

C. Removes the first node that contains the given value from the list

D. Reverses the linked list

Solution

1. Step 1: Understand the operation

   Deleting a node by value means searching for the first node that matches the value and removing it from the linked list.

2. Step 2: Compare options

   Only Removes the first node that contains the given value from the list describes removing the first node with the given value. Others describe adding, finding without removing, or reversing, which are different operations.

3. Final Answer:

   Removes the first node that contains the given value from the list -> Option C

4. Quick Check:

   Delete Node by Value = Remove first matching node [OK]

Hint: Deleting by value removes first matching node only [OK]

Common Mistakes:

• Confusing deletion by value with adding a node
• Thinking it removes all nodes with that value
• Assuming it reverses the list

2. Which of the following is the correct C# syntax to delete a node by value from a singly linked list?

easy

A. void DeleteNodeByValue(int value) { /* code */ }

B. public void DeleteNode(int value) { /* code */ }

C. public int DeleteNode(int value) { /* code */ }

D. DeleteNode(int value) => { /* code */ }

Solution

1. Step 1: Check method signature conventions in C#

   In C#, methods usually have an access modifier like 'public' and a return type. Deleting a node usually returns void.

2. Step 2: Analyze options

   public void DeleteNode(int value) { /* code */ } uses 'public void DeleteNode(int value)', which is correct. void DeleteNodeByValue(int value) { /* code */ } misses access modifier. public int DeleteNode(int value) { /* code */ } returns int which is uncommon here. DeleteNode(int value) => { /* code */ } uses invalid lambda syntax for method.

3. Final Answer:

   public void DeleteNode(int value) { /* code */ } -> Option B

4. Quick Check:

   Correct method signature = public void DeleteNode(int value) { /* code */ } [OK]

Hint: Use public void method with parameter for deletion [OK]

Common Mistakes:

• Omitting access modifier
• Using wrong return type
• Incorrect method syntax

3. Given the linked list: 10 -> 20 -> 30 -> 40 -> null, what is the list after deleting the node with value 30 using the DeleteNodeByValue method?

medium

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

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

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

D. 10 -> 30 -> 40 -> null

Solution

1. Step 1: Identify the node to delete

   The node with value 30 is the third node in the list.

2. Step 2: Remove the node and link previous to next

   After deleting 30, the node with value 20 points directly to 40, so the list becomes 10 -> 20 -> 40 -> null.

3. Final Answer:

   10 -> 20 -> 40 -> null -> Option A

4. Quick Check:

   Deleted 30, list skips it [OK]

Hint: Remove node and link previous to next node [OK]

Common Mistakes:

• Not updating links after deletion
• Deleting wrong node
• Leaving deleted node in list

4. Consider this code snippet to delete a node by value in a singly linked list:

public void DeleteNode(int value) {
  Node current = head;
  Node previous = null;
  while (current != null && current.data != value) {
    previous = current;
    current = current.next;
  }
  if (current == null) return;
  previous.next = current.next;
}

What error will occur when deleting the head node?

medium

A. Compilation error due to missing return type

B. No error, deletion works fine

C. Infinite loop because current never changes

D. NullReferenceException because previous is null when deleting head

Solution

1. Step 1: Analyze deletion of head node

   If the node to delete is head, previous is still null because loop never runs.

2. Step 2: Check assignment previous.next = current.next

   Since previous is null, trying to access previous.next causes NullReferenceException.

3. Final Answer:

   NullReferenceException because previous is null when deleting head -> Option D

4. Quick Check:

   Deleting head needs special case to avoid null error [OK]

Hint: Handle head deletion separately to avoid null errors [OK]

Common Mistakes:

• Not checking if node to delete is head
• Assuming previous is never null
• Ignoring null pointer exceptions

5. You have a singly linked list: 5 -> 10 -> 15 -> 20 -> 25 -> null. You want to delete the node with value 5 (head), then delete the node with value 20. What will the list look like after these two deletions?

hard

A. 10 -> 15 -> 25 -> null

B. 5 -> 10 -> 15 -> 25 -> null

C. 10 -> 15 -> 20 -> 25 -> null

D. 15 -> 25 -> null

Solution

1. Step 1: Delete head node with value 5

   Removing head means head moves to next node, so list becomes 10 -> 15 -> 20 -> 25 -> null.

2. Step 2: Delete node with value 20

   Find node 20 and remove it by linking 15 to 25, resulting in 10 -> 15 -> 25 -> null.

3. Final Answer:

   10 -> 15 -> 25 -> null -> Option A

4. Quick Check:

   Deleted head and middle node correctly [OK]

Hint: Delete head first, then link previous to next for others [OK]

Common Mistakes:

• Not updating head after deleting first node
• Not linking nodes correctly after deletion
• Deleting wrong nodes