Pop Using Linked List Node in DSA Python - Execution Trace

Concept Flow - Pop Using Linked List Node

Start with head node

↓

Check if list is empty?

↓

Return None

↓

Move head to next node

↓

Disconnect popped node

↓

Return popped node's value

This flow shows how to remove the first node from a linked list by updating the head pointer and returning the removed node's value.

Execution Sample

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def pop(head):
    if head is None:
        return None, None
    popped = head
    head = head.next
    popped.next = None
    return popped.val, head

This code removes the first node from a linked list and returns its value.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Start with list: 10 -> 20 -> 30 -> None	10 -> 20 -> 30 -> None	head points to node with val=10	head -> [10] -> [20] -> [30] -> None
2	Check if list is empty	10 -> 20 -> 30 -> None	head is not None	head -> [10] -> [20] -> [30] -> None
3	Save head node to popped	10 -> 20 -> 30 -> None	popped = head (val=10)	popped -> [10] -> [20] -> [30] -> None
4	Move head to next node	10 -> 20 -> 30 -> None	head = head.next (val=20)	head -> [20] -> [30] -> None
5	Disconnect popped node	10 (popped) 20 -> 30 -> None	popped.next = None	popped -> [10] -> None head -> [20] -> [30] -> None
6	Return popped node's value	20 -> 30 -> None	Return 10	head -> [20] -> [30] -> None

💡 After popping, head points to second node; popped node is disconnected and returned.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	Final
head	Node(10)	Node(10)	Node(20)	Node(20)	Node(20)
popped	None	Node(10)	Node(10)	Node(10)	Node(10)
popped.next	N/A	Node(20)	Node(20)	None	None

Key Moments - 3 Insights

Why do we set popped.next = None after moving head?

Setting popped.next = None disconnects the popped node from the list, preventing accidental links. See step 5 in execution_table where this happens.

What happens if the list is empty (head is None)?

The function returns None immediately without changes, as shown in step 2 where the check for empty list occurs.

Why do we update head to head.next before disconnecting popped?

We update head first to keep the list connected starting from the second node, then disconnect popped to isolate it. This order is shown in steps 4 and 5.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of head after step 4?

A Node with value 10

B Node with value 30

C Node with value 20

D None

💡 Hint

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

At which step is the popped node disconnected from the list?

A Step 5

B Step 4

C Step 3

D Step 6

💡 Hint

Look for 'popped.next = None' in the 'Operation' column of execution_table.

If the list was empty, what would the pop function return?

A The value of the head node

B None

C An error

D The last node's value

💡 Hint

Refer to step 2 in execution_table where the empty list check happens.

Concept Snapshot

Pop Using Linked List Node:
- Check if head is None; if yes, return None
- Save head node as popped
- Move head to next node
- Disconnect popped node by setting popped.next = None
- Return popped node's value
This removes the first node and updates the list head.

Full Transcript

This concept shows how to remove the first node from a linked list. We start by checking if the list is empty. If it is, we return None. Otherwise, we save the current head node as the popped node. Then, we move the head pointer to the next node in the list. After that, we disconnect the popped node by setting its next pointer to None. Finally, we return the value of the popped node. This process updates the linked list by removing its first element.