Pop Using Linked List Node in DSA Java - Execution Trace

Concept Flow - Pop Using Linked List Node

Check if list is empty

No↓

Store head node data

↓

Move head pointer to next node

↓

Return stored data

Yes↓

Return null or error

The pop operation removes the first node (head) from the linked list by moving the head pointer to the next node and returning the removed node's data.

Execution Sample

DSA Java

class Node {
  int data;
  Node next;
}

Node head;

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

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

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Initial list	10 -> 20 -> 30 -> null	head points to node with 10	head -> [10] -> [20] -> [30] -> null
2	Check if head is null	10 -> 20 -> 30 -> null	head unchanged	head -> [10] -> [20] -> [30] -> null
3	Store head data	10 -> 20 -> 30 -> null	head unchanged	head -> [10] -> [20] -> [30] -> null
4	Move head to next	20 -> 30 -> null	head now points to node with 20	head -> [20] -> [30] -> null
5	Return stored data	20 -> 30 -> null	head unchanged	head -> [20] -> [30] -> null
6	Pop complete	20 -> 30 -> null	head points to 20	head -> [20] -> [30] -> null
7	Pop again	20 -> 30 -> null	head points to 20	head -> [20] -> [30] -> null
8	Store head data	20 -> 30 -> null	head unchanged	head -> [20] -> [30] -> null
9	Move head to next	30 -> null	head now points to node with 30	head -> [30] -> null
10	Return stored data	30 -> null	head unchanged	head -> [30] -> null
11	Pop again	30 -> null	head points to 30	head -> [30] -> null
12	Store head data	30 -> null	head unchanged	head -> [30] -> null
13	Move head to next	null	head now points to null	head -> null
14	Return stored data	null	head unchanged	head -> null
15	Pop on empty list	null	head is null	head -> null
16	Return -1 or error	null	head unchanged	head -> null

💡 Execution stops when head becomes null, indicating the list is empty and no more nodes to pop.

Variable Tracker

Variable	Start	After Step 4	After Step 9	After Step 13	Final
head	points to node with 10	points to node with 20	points to node with 30	null	null
stored data	undefined	10	20	30	undefined

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 remove. This is shown in execution_table row 2 and row 15 where the check prevents errors.

What happens to the removed node after pop?

The removed node is no longer referenced by head, so it is effectively removed from the list. This is shown in execution_table rows 4 and 13 where head moves to the next node.

Why do we store head data before moving head pointer?

Because once head moves to next, the original node is lost. Storing data first ensures we can return the popped value. See execution_table rows 3, 8, and 12.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what does the head pointer point to?

A Node with data 10

B Node with data 30

C Node with data 20

D null

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns at step 4.

At which step does the linked list become empty?

A Step 9

B Step 13

C Step 15

D Step 16

💡 Hint

Look at the 'Nodes in List' and 'Pointer Changes' columns to see when head points to null.

If the list is empty, what value does pop return according to the execution?

A -1

B 0

C The data of the last node

D null

💡 Hint

See step 16 where pop returns -1 when head is null.

Concept Snapshot

Pop removes the first node (head) from a linked list.
Check if head is null (empty list) before popping.
Store head data, move head to next node.
Return stored data.
If empty, return -1 or error.
Head pointer always moves forward.

Full Transcript

Pop operation on a linked list removes the first node by moving the head pointer to the next node and returning the removed node's data. Initially, the list has nodes 10, 20, 30 with head pointing to 10. We check if the list is empty by verifying if head is null. If not empty, we store the data of the head node, move head to the next node, and return the stored data. This process repeats until the list becomes empty, at which point pop returns -1. The execution table shows each step, pointer changes, and the visual state of the list. Variable tracker shows how head and stored data change after each pop. Key moments clarify why we check for empty list, why we store data before moving head, and what happens to removed nodes. The visual quiz tests understanding of pointer changes, empty list detection, and return values on empty pop.

Practice

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

easy

A. Removes and returns the first node of the linked list

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

C. Removes the last node without returning it

D. Returns the size of the linked list

Solution

1. Step 1: Understand pop operation in linked list context

   The pop operation removes the top element, which is the first node in a linked list acting like a stack.

2. Step 2: Identify what is returned

   Pop returns the value of the removed first node and updates the head pointer to the next node.

3. Final Answer:

   Removes and returns the first node of the linked list -> Option A

4. Quick Check:

   Pop = Remove first node [OK]

Hint: Pop removes the head node and returns it [OK]

Common Mistakes:

• Thinking pop removes the last node
• Confusing pop with push operation
• Assuming pop returns list size

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

easy

A. head = head.next; return head.data;

B. Node temp = head; head = head.next; return temp.data;

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

D. return head.next.data; head = head.next;

Solution

1. Step 1: Check correct order of operations

   We must save the current head node in a temp variable before moving head to next.

2. Step 2: Return the data of the removed node

   Return the data from the saved temp node, not the new head.

3. Final Answer:

   Node temp = head; head = head.next; return temp.data; -> Option B

4. Quick Check:

   Save head, move head, return old data [OK]

Hint: Save head before moving it to next [OK]

Common Mistakes:

• Returning data from new head instead of old
• Not saving old head before moving
• Returning data before updating head

3. Given the linked list nodes: 10 -> 20 -> 30 -> null, what is the list after one pop operation using this code?

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

medium

A. 30 -> null

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

C. 20 -> 30 -> null

D. null

Solution

1. Step 1: Identify the node removed by pop()

   Pop removes the head node which has value 10.

2. Step 2: Update head to next node

   Head moves from 10 to 20, so list becomes 20 -> 30 -> null.

3. Final Answer:

   20 -> 30 -> null -> Option C

4. Quick Check:

   Pop removes first node 10 [OK]

Hint: Pop removes first node, head moves next [OK]

Common Mistakes:

• Assuming pop removes last node
• Not updating head pointer
• Returning old list unchanged

4. What is the error in this pop method for a singly linked list?

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

medium

A. It skips the first node and removes the second node instead

B. It causes a null pointer exception when list is empty

C. It returns the old head node instead of new head

D. It does not update the head pointer

Solution

1. Step 1: Analyze which node is assigned to temp

   Temp is assigned head.next, so the first node is skipped.

2. Step 2: Understand effect on head and return

   Head moves to second node, so first node is never removed; second node is removed instead.

3. Final Answer:

   It skips the first node and removes the second node instead -> Option A

4. Quick Check:

   Pop must remove first node, not second [OK]

Hint: Pop must save head, not head.next [OK]

Common Mistakes:

• Assigning temp to head.next instead of head
• Returning the new head instead of the old head
• Returning wrong node

5. You have a linked list: 5 -> 15 -> 25 -> 35 -> null. You perform two pop operations using this method:

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

What is the state of the list after these two pops?

hard

A. null

B. 15 -> 25 -> 35 -> null

C. 35 -> null

D. 25 -> 35 -> null

Solution

1. Step 1: First pop removes 5

   After first pop, head moves from 5 to 15, list is 15 -> 25 -> 35 -> null.

2. Step 2: Second pop removes 15

   After second pop, head moves from 15 to 25, list is 25 -> 35 -> null.

3. Final Answer:

   25 -> 35 -> null -> Option D

4. Quick Check:

   Two pops remove first two nodes [OK]

Hint: Each pop removes head, moves head next [OK]

Common Mistakes:

• Removing last nodes instead of first
• Not updating head after each pop
• Confusing pop with push