Pop Using Linked List Node in DSA Python

Mental Model

Pop means removing the last item from a linked list and returning it. We find the second last node and cut off the last node.

Analogy: Imagine a chain of paper clips linked together. To remove the last clip, you hold the second last clip and unhook the last one.

head -> 1 -> 2 -> 3 -> null
                 ↑
               tail

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, pop last node

Goal: Remove the last node (3) and update the list to end at node 2

Step 1: start at head node (1)

head -> [1] -> 2 -> 3 -> null

Why: We need to find the second last node to remove the last

Step 2: move to next node (2)

head -> 1 -> [2] -> 3 -> null

Why: Check if next node is last; if yes, we can pop

Step 3: next node (3) is last, so set node 2's next to null

head -> 1 -> 2 -> null    popped node: 3

Why: Cut off last node by removing reference from second last

Result:

head -> 1 -> 2 -> null
popped node value: 3

Annotated Code

DSA Python

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

class LinkedList:
    def __init__(self):
        self.head = None

    def append(self, val):
        new_node = Node(val)
        if not self.head:
            self.head = new_node
            return
        curr = self.head
        while curr.next:
            curr = curr.next
        curr.next = new_node

    def pop(self):
        if not self.head:
            return None  # empty list
        if not self.head.next:
            val = self.head.val
            self.head = None
            return val
        curr = self.head
        while curr.next.next:
            curr = curr.next  # advance to second last node
        val = curr.next.val
        curr.next = None  # remove last node
        return val

    def __str__(self):
        vals = []
        curr = self.head
        while curr:
            vals.append(str(curr.val))
            curr = curr.next
        return ' -> '.join(vals) + ' -> null'

# Driver code
ll = LinkedList()
ll.append(1)
ll.append(2)
ll.append(3)
print("Before pop:", ll)
popped_value = ll.pop()
print("Popped node value:", popped_value)
print("After pop:", ll)

if not self.head: return None # empty list

handle empty list case by returning None

if not self.head.next: val = self.head.val self.head = None return val

handle single node list by removing head and returning its value

while curr.next.next: curr = curr.next # advance to second last node

move curr to second last node to prepare for pop

val = curr.next.val curr.next = None # remove last node return val

store last node value, remove last node by cutting link, return value

Output

Before pop: 1 -> 2 -> 3 -> null

Popped node value: 3

After pop: 1 -> 2 -> null

Complexity Analysis

Time: O(n) because we traverse the list once to find the second last node

Space: O(1) because we only use a few pointers, no extra space grows with input

vs Alternative: Compared to arrays where pop from end is O(1), linked list pop requires traversal O(n) because no direct access to last node

Edge Cases

empty list

pop returns None without error

DSA Python

if not self.head:
    return None  # empty list

single node list

pop removes the only node and sets head to None

DSA Python

if not self.head.next:
    val = self.head.val
    self.head = None
    return val

When to Use This Pattern

When you need to remove the last element from a singly linked list, use the pop pattern by finding the second last node and cutting off the last.

Common Mistakes

Mistake: Trying to pop by moving to the last node directly and setting it to None
Fix: Move to the second last node and set its next pointer to None to remove last node

Mistake: Not handling empty or single node list cases causing errors
Fix: Add checks for empty list and single node list before traversal

Summary

Removes the last node from a singly linked list and returns its value.

Use when you want to remove the tail element but only have access to head.

You must find the second last node to cut off the last node since singly linked lists have no backward pointers.

Practice

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

easy

A. Removes the last node and returns its value

B. Adds a new node at the beginning

C. Searches for a node with a specific value

D. Counts the total number of nodes

Solution

1. Step 1: Understand pop operation in linked list

   Pop means removing the last element from the list and returning its value.

2. Step 2: Identify pop behavior with nodes

   In a linked list, pop removes the tail node and returns its data.

3. Final Answer:

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

4. Quick Check:

   Pop = Remove last node [OK]

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

Common Mistakes:

• Confusing pop with push (adding nodes)
• Thinking pop removes the first node
• Assuming pop searches for a value

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

easy

A. def pop(self):\n if not self.head:\n return None\n current = self.head\n while current.next.next:\n current = current.next\n popped = current.next\n current.next = None\n return popped.data

B. def pop(self):\n popped = self.head\n self.head = self.head.next\n return popped.data

C. def pop(self):\n return self.head.data

D. def pop(self):\n self.head = None

Solution

1. Step 1: Check pop logic for singly linked list

   Pop removes the last node, so we must traverse to the second last node.

2. Step 2: Verify syntax correctness

   def pop(self):\n if not self.head:\n return None\n current = self.head\n while current.next.next:\n current = current.next\n popped = current.next\n current.next = None\n return popped.data correctly traverses until current.next.next is None, then removes last node and returns its data.

3. Final Answer:

   Code that traverses to second last node -> Option A

4. Quick Check:

   Traverse to second last node, remove last [OK]

Hint: Pop needs traversal to second last node before removal [OK]

Common Mistakes:

• Removing head instead of last node
• Not checking for empty list
• Returning node instead of node data

3. What will be the output after popping a node from the linked list 1 -> 2 -> 3 -> null using the pop method below?

def pop(self):
    if not self.head:
        return None
    if not self.head.next:
        val = self.head.data
        self.head = None
        return val
    current = self.head
    while current.next.next:
        current = current.next
    val = current.next.data
    current.next = None
    return val

medium

A. 2 -> 3 -> null

B. 1 -> 3 -> null

C. 1 -> 2 -> null

D. 3 -> null

Solution

1. Step 1: Identify the node popped

   Pop removes the last node, which is 3 in the list 1 -> 2 -> 3 -> null.

2. Step 2: Determine list after pop

   After removing 3, the list becomes 1 -> 2 -> null.

3. Final Answer:

   1 -> 2 -> null -> Option C

4. Quick Check:

   Last node removed, list ends at 2 [OK]

Hint: Pop removes last node, list shortens by one [OK]

Common Mistakes:

• Removing first node instead of last
• Not updating next pointer to None
• Returning wrong list after pop

4. Find the error in this pop method for a singly linked list with at least two nodes:

def pop(self):
    current = self.head
    while current.next:
        current = current.next
    val = current.data
    current = None
    return val

medium

A. It does not handle empty list case

B. It does not update the previous node's next pointer

C. It returns None instead of node data

D. It removes the head instead of last node

Solution

1. Step 1: Analyze traversal and removal

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

2. Step 2: Check pointer update

   Setting current = None only changes local variable, does not update previous node's next pointer to None, so last node remains linked.

3. Final Answer:

   It does not update the previous node's next pointer -> Option B

4. Quick Check:

   Must update previous.next to None to remove last node [OK]

Hint: Always update previous node's next to None when popping last node [OK]

Common Mistakes:

• Not handling empty list
• Not tracking previous node
• Setting local variable to None instead of pointer

5. Given a linked list with nodes 10 -> 20 -> 30 -> 40 -> null, write the sequence of node values after two consecutive pop operations using a correct pop method that removes the last node each time.

hard

A. 10 -> 30 -> null

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

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

D. 10 -> 20 -> null

Solution

1. Step 1: First pop removes last node 40

   List changes from 10 -> 20 -> 30 -> 40 -> null to 10 -> 20 -> 30 -> null.

2. Step 2: Second pop removes new last node 30

   List changes from 10 -> 20 -> 30 -> null to 10 -> 20 -> null.

3. Final Answer:

   10 -> 20 -> null -> Option D

4. Quick Check:

   Two pops remove last two nodes [OK]

Hint: Each pop removes last node, shorten list stepwise [OK]

Common Mistakes:

• Removing nodes from start instead of end
• Not updating list after first pop
• Confusing node values after pops