Pop Using Linked List Node in DSA Javascript - Execution Trace

Concept Flow - Pop Using Linked List Node

Start at head

↓

Is head null?

Yes→List empty, nothing to pop

No↓

Is head.next null?

Yes→List has one node, pop that node

No↓

Traverse to second last node

↓

Set second last node.next = null

↓

Return popped node

↓

Done

Start from the head node, check if list is empty or has one node, otherwise traverse to second last node, remove last node by setting next to null, and return popped node.

Execution Sample

DSA Javascript

function pop(head) {
  if (!head) return null;
  if (!head.next) {
    let popped = head;
    head = null;
    return popped;
  }
  let current = head;
  while (current.next.next) {
    current = current.next;
  }
  let popped = current.next;
  current.next = null;
  return popped;
}

This code removes the last node from a singly linked list and returns it.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Check if head is null	1 -> 2 -> 3 -> null	head points to node 1	1 -> 2 -> 3 -> null
2	Check if head.next is null	1 -> 2 -> 3 -> null	head.next points to node 2	1 -> 2 -> 3 -> null
3	Initialize current = head	1 -> 2 -> 3 -> null	current points to node 1	1 -> 2 -> 3 -> null
4	Traverse while current.next.next exists	1 -> 2 -> 3 -> null	current moves to node 2	1 -> 2 -> 3 -> null
5	Set popped = current.next (node 3)	1 -> 2 -> 3 -> null	popped points to node 3	1 -> 2 -> 3 -> null
6	Set current.next = null (remove last node)	1 -> 2 -> null	node 2.next set to null	1 -> 2 -> null
7	Return popped node (node 3)	1 -> 2 -> null	popped node 3 returned	1 -> 2 -> null
8	End	1 -> 2 -> null	No pointer changes	1 -> 2 -> null

💡 Traversal stops when current.next.next is null, meaning current is second last node.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	After Step 6	Final
head	node 1	node 1	node 1	node 1	node 1	node 1
current	undefined	node 1	node 2	node 2	node 2	node 2
popped	undefined	undefined	undefined	node 3	node 3	node 3

Key Moments - 3 Insights

Why do we check current.next.next instead of current.next in the loop?

We check current.next.next to stop at the second last node, so we can remove the last node by setting current.next to null. See step 4 in execution_table.

What happens if the list has only one node?

If head.next is null, it means only one node exists. The code pops that single node by returning it and setting head to null, effectively emptying the list.

Why do we set current.next = null to pop the last node?

Setting current.next = null removes the last node from the list by breaking the link. The last node becomes unreachable, effectively popped. See step 6.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, which node does 'current' point to?

A Node 1

B Node 3

C Node 2

D Null

💡 Hint

Check the 'Pointer Changes' column at step 4 in execution_table.

At which step does the last node get removed from the list?

A Step 5

B Step 6

C Step 7

D Step 8

💡 Hint

Look for the step where current.next is set to null in execution_table.

If the list had only one node, what would the function return?

A The single node

B Null

C The head's next node

D An error

💡 Hint

Refer to the updated code and explanation for single node case.

Concept Snapshot

Pop Using Linked List Node:
- Start at head
- If list empty, return null
- If list has one node, pop that node and set head to null
- Traverse to second last node
- Set second last node.next = null
- Return popped last node
- Removes last node from singly linked list

Full Transcript

This visualization shows how to pop the last node from a singly linked list. We start by checking if the list is empty or has only one node. If empty, we return null because there's no node to pop. If only one node exists, we pop that node by returning it and setting head to null, effectively emptying the list. Otherwise, we traverse the list to find the second last node by checking current.next.next. Once found, we save the last node as popped, then remove it by setting current.next to null. Finally, we return the popped node. The execution table tracks each step, pointer changes, and the list's visual state. Key moments clarify why we check current.next.next and how the last node is removed. The quiz tests understanding of pointer positions and edge cases. This method safely removes the last node from a singly linked list.

Practice

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

easy

A. Adds a new node at the end

B. Returns the value of the last node without removing it

C. Removes the first node and returns its value

D. Removes the last node and returns its value

Solution

1. Step 1: Understand pop operation in linked list

   The pop operation removes the last node from the linked list and returns its value.

2. Step 2: Differentiate from other operations

   Adding a node is push, removing first node is shift, and peeking returns value without removal.

3. Final Answer:

   Removes the last node and returns its value -> Option D

4. Quick Check:

   pop() removes last node = B [OK]

Hint: Pop always removes the last node in linked list [OK]

Common Mistakes:

• Confusing pop with push (adding nodes)
• Thinking pop removes the first node
• Assuming pop only returns value without removal

2. Which of the following is the correct syntax to remove the last node in a singly linked list in JavaScript?

easy

A. list.popNode()

B. list.deleteTail()

C. list.pop()

D. list.removeLast()

Solution

1. Step 1: Identify standard method name

   In JavaScript linked list implementations, pop() is the standard method to remove the last node.

2. Step 2: Check other options

   Other method names like popNode(), removeLast(), or deleteTail() are not standard or common.

3. Final Answer:

   list.pop() -> Option C

4. Quick Check:

   pop() removes last node syntax = D [OK]

Hint: Use pop() to remove last node in linked list [OK]

Common Mistakes:

• Using non-standard method names
• Confusing pop() with other removal methods
• Assuming popNode() is built-in

3. Consider the following linked list and pop operation in JavaScript:

class Node {
  constructor(value) {
    this.value = value;
    this.next = null;
  }
}

class LinkedList {
  constructor() {
    this.head = null;
  }
  push(value) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }
  pop() {
    if (!this.head) return null;
    if (!this.head.next) {
      const val = this.head.value;
      this.head = null;
      return val;
    }
    let current = this.head;
    while (current.next.next) {
      current = current.next;
    }
    const val = current.next.value;
    current.next = null;
    return val;
  }
}

const list = new LinkedList();
list.push(10);
list.push(20);
list.push(30);
const poppedValue = list.pop();
console.log(poppedValue);

What will be printed?

medium

A. 30

B. 20

C. 10

D. null

Solution

1. Step 1: Trace push operations

   Nodes added in order: 10 -> 20 -> 30

2. Step 2: Trace pop operation

   Pop removes last node (30) and returns its value.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   Last node value popped = 30 [OK]

Hint: Pop returns last pushed value in linked list [OK]

Common Mistakes:

• Confusing pop with removing first node
• Expecting 20 instead of 30
• Assuming pop returns null if list not empty

4. The following pop() method for a singly linked list has a bug. What is the error?

pop() {
  if (!this.head) return null;
  let current = this.head;
  while (current.next) {
    current = current.next;
  }
  const val = current.value;
  current = null;
  return val;
}

medium

A. It does not update the previous node's next to null

B. It does not handle empty list

C. It returns the wrong node's value

D. It causes infinite loop

Solution

1. Step 1: Analyze traversal to last node

   The code moves current to last node correctly but does not track previous node.

2. Step 2: Check removal logic

   Setting current = null only changes local variable, does not remove last node from list.

3. Step 3: Identify missing update

   Previous node's next should be set to null to remove last node.

4. Final Answer:

   It does not update the previous node's next to null -> Option A

5. Quick Check:

   Missing previous.next = null causes bug = A [OK]

Hint: Update previous node's next to null to remove last node [OK]

Common Mistakes:

• Not tracking previous node during traversal
• Setting current = null does not remove node
• Ignoring empty list check

5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> 35 -> null.
You perform two pop operations in a row. What is the state of the linked list after these operations?

hard

A. 15 -> 25 -> 35 -> null

B. 5 -> 15 -> null

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

D. null

Solution

1. Step 1: Initial list nodes

   List is 5 -> 15 -> 25 -> 35 -> null.

2. Step 2: First pop removes last node

   Pop removes 35, list becomes 5 -> 15 -> 25 -> null.

3. Step 3: Second pop removes new last node

   Pop removes 25, list becomes 5 -> 15 -> null.

4. Final Answer:

   5 -> 15 -> null -> Option B

5. Quick Check:

   Two pops remove last two nodes = A [OK]

Hint: Each pop removes last node, so list shrinks from tail [OK]

Common Mistakes:

• Removing nodes from the front instead of end
• Removing only one node after two pops
• Assuming list becomes empty after two pops