Pop Using Linked List Node in DSA C# - Execution Trace

Concept Flow - Pop Using Linked List Node

Check if list is empty

No↓

Save head node value

↓

Move head pointer to next node

↓

Return saved value

↓

Done

Yes↓

Return error or null

The pop operation removes the first node (head) from the linked list, returns its value, and updates the head pointer to the next node.

Execution Sample

DSA C#

int Pop()
{
  if (head == null) return -1;
  int val = head.value;
  head = head.next;
  return val;
}

This code removes the first node from the linked list and returns its value, or -1 if the list is empty.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Check if head is null	10 -> 20 -> 30 -> null	head points to node with value 10	10 -> 20 -> 30 -> null
2	Save head node value	10 -> 20 -> 30 -> null	head points to node with value 10	10 -> 20 -> 30 -> null
3	Move head to next node	20 -> 30 -> null	head now points to node with value 20	20 -> 30 -> null
4	Return saved value	20 -> 30 -> null	head points to node with value 20	20 -> 30 -> null
5	End of pop operation	20 -> 30 -> null	head points to node with value 20	20 -> 30 -> null

💡 Pop ends after removing the head node and updating the head pointer.

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
head	points to node 10	points to node 10	points to node 20	points to node 20
val	undefined	10	10	10
list size	3 nodes	3 nodes	2 nodes	2 nodes

Key Moments - 3 Insights

Why do we check if head is null before popping?

Because if head is null, the list is empty and there is no node to pop. This is shown in execution_table step 1 where we check head before proceeding.

Why do we update head to head.next after saving the value?

Updating head to head.next removes the first node from the list logically. The saved value is kept to return after the head moves, as shown in steps 2 and 3.

What happens to the removed node after pop?

The removed node is no longer referenced by head, so it can be cleaned up by memory management. The list visually no longer includes it after step 3.

Visual Quiz - 3 Questions

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

A 20

B 10

C null

D -1

💡 Hint

Check the 'val' variable in variable_tracker after step 2.

At which step does the head pointer move to the next node?

A Step 3

B Step 1

C Step 2

D Step 4

💡 Hint

Look at the 'Pointer Changes' column in execution_table.

If the list was empty (head == null), what would the pop function return?

A The value of the head node

B null

C -1

D 0

💡 Hint

Refer to the code in execution_sample and the check in execution_table step 1.

Concept Snapshot

Pop removes the first node from a linked list.
Check if list is empty (head == null).
Save head node's value.
Move head pointer to next node.
Return saved value.
List size decreases by one.

Full Transcript

Pop Using Linked List Node removes the first node from the list. First, it checks if the list is empty by seeing if head is null. If empty, it returns -1. Otherwise, it saves the value of the head node. Then it moves the head pointer to the next node, effectively removing the first node. Finally, it returns the saved value. This operation reduces the list size by one and updates the head pointer to the new first node.

Practice

1. What does the pop operation do in a linked list when implemented using nodes?

easy

A. Searches for a node with a specific value

B. Adds a new node at the end of the linked list

C. Counts the total number of nodes in the linked list

D. Removes and returns the first node from the linked list

Solution

1. Step 1: Understand the pop operation

   The pop operation removes an element from a data structure. In a linked list, it usually removes the first node (head).

2. Step 2: Identify the effect on the linked list

   Removing the first node means the head pointer moves to the next node, and the removed node is returned.

3. Final Answer:

   Removes and returns the first node from the linked list -> Option D

4. Quick Check:

   Pop = Remove first node [OK]

Hint: Pop removes the first node in a linked list [OK]

Common Mistakes:

• Confusing pop with adding nodes
• Thinking pop removes the last node
• Assuming pop counts nodes

2. Which of the following is the correct syntax to pop a node from the head of a singly linked list in C#?

easy

A. head = null; return head;

B. head = head.next; return head;

C. Node temp = head; head = head.next; return temp;

D. Node temp = head.next; head = temp; return head;

Solution

1. Step 1: Save the current head node

   We store the current head in a temporary variable to return it later.

2. Step 2: Move head to the next node

   Update the head pointer to the next node in the list.

3. Step 3: Return the removed node

   Return the saved node which was the original head.

4. Final Answer:

   Node temp = head; head = head.next; return temp; -> Option C

5. Quick Check:

   Save head, move head next, return saved [OK]

Hint: Save head, move head to next, return saved node [OK]

Common Mistakes:

• Returning the new head instead of removed node
• Not saving the original head before moving
• Setting head to null incorrectly

3. Consider the following C# code for popping a node from a linked list:

Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);

Node popped = Pop(ref head);
Console.WriteLine(popped.data);

What will be printed?

Assume Pop removes the first node and updates head.

medium

A. 10

B. 20

C. 30

D. NullReferenceException

Solution

1. Step 1: Understand initial linked list

   The list has nodes 10 -> 20 -> 30. Head points to node with data 10.

2. Step 2: Pop removes the first node

   Pop removes node 10 and updates head to node 20.

3. Step 3: Output the popped node's data

   The popped node's data is 10, so Console.WriteLine prints 10.

4. Final Answer:

   10 -> Option A

5. Quick Check:

   First node data popped = 10 [OK]

Hint: Pop returns first node's data before head moves [OK]

Common Mistakes:

• Printing head data after pop instead of popped node
• Confusing node values order
• Assuming pop returns null

4. The following C# method is intended to pop the first node from a linked list:

Node Pop(Node head) {
    Node temp = head;
    head = head.next;
    return temp;
}

What is the main problem with this code?

medium

A. It does not update the original head outside the method

B. It returns the wrong node

C. It causes a null pointer exception always

D. It deletes the entire linked list

Solution

1. Step 1: Analyze parameter passing

   The method receives head by value, so changes to head inside do not affect the caller's head.

2. Step 2: Understand effect on original list

   Since head is not updated outside, the original list's head remains unchanged, so pop does not remove the node effectively.

3. Final Answer:

   It does not update the original head outside the method -> Option A

4. Quick Check:

   Head passed by value, original not updated [OK]

Hint: Use ref to update head outside method [OK]

Common Mistakes:

• Assuming head changes inside method affect caller
• Thinking it returns wrong node
• Believing it causes runtime error

5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> null.
You want to pop two nodes using a method that updates the head by reference.
After popping twice, what is the state of the linked list?

hard

A. 15 -> 25 -> null

B. 25 -> null

C. 5 -> 15 -> 25 -> null

D. null

Solution

1. Step 1: Initial linked list nodes

   The list starts as 5 -> 15 -> 25 -> null, head points to node with 5.

2. Step 2: Pop first node

   Pop removes node 5, head moves to node 15, list becomes 15 -> 25 -> null.

3. Step 3: Pop second node

   Pop removes node 15, head moves to node 25, list becomes 25 -> null.

4. Final Answer:

   25 -> null -> Option B

5. Quick Check:

   Two pops remove first two nodes [OK]

Hint: Each pop removes head, so two pops leave third node [OK]

Common Mistakes:

• Not updating head after pop
• Assuming popped nodes remain in list
• Confusing node order after pops