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 data

↓

Move head pointer to next node

↓

Delete old head node

↓

Return saved data

↓

Done

Yes↓

Return error or null

This flow shows how to remove the first node from a linked list by saving its data, moving the head pointer, deleting the old node, and returning the data.

Execution Sample

DSA C++

struct Node {
  int data;
  Node* next;
};

int pop(Node*& head) {
  if (!head) return -1;
  int val = head->data;
  Node* temp = head;
  head = head->next;
  delete temp;
  return val;
}

This code removes the first node from the linked list, returns its data, and updates the head pointer.

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 10	10 -> 20 -> 30 -> null
2	Save head data (10)	10 -> 20 -> 30 -> null	head points to node with 10	10 -> 20 -> 30 -> null
3	Move head to next node	10 -> 20 -> 30 -> null	head now points to node with 20	20 -> 30 -> null
4	Delete old head node (10)	20 -> 30 -> null	head points to node with 20	20 -> 30 -> null
5	Return saved data (10)	20 -> 30 -> null	head points to node with 20	20 -> 30 -> null
6	Check if head is null (empty list case)	null	head is null	null
7	Return -1 for empty list	null	head is null	null

💡 Execution stops after returning the popped value or -1 if the list is empty.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
head	points to node with 10	points to node with 10	points to node with 20	points to node with 20	points to node with 20
temp	undefined	points to node with 10	points to node with 10	deleted	deleted
val	undefined	10	10	10	10

Key Moments - 3 Insights

Why do we save the head node's data before moving the head pointer?

Because after moving the head pointer to the next node, the original head node is deleted and its data would be lost. See execution_table step 2 and 3.

What happens if the list is empty when we try to pop?

The function checks if head is null at step 1 and step 6. If it is null, it returns -1 immediately without changing anything.

Why do we delete the old head node after moving the head pointer?

To free memory and avoid memory leaks. The old head node is no longer part of the list after head moves. See execution_table step 4.

Visual Quiz - 3 Questions

Look at the execution_table, what does the head pointer point to after step 3?

A Node with data 20

B Node with data 30

C Node with data 10

D Null

💡 Hint

Check the 'Pointer Changes' column at step 3 in execution_table.

At which step is the old head node deleted?

A Step 2

B Step 4

C Step 3

D Step 5

💡 Hint

Look at the 'Operation' column for deletion in execution_table.

If the list is empty, what value does the pop function return?

A null

B 0

C -1

D Undefined

💡 Hint

See execution_table steps 6 and 7 for empty list handling.

Concept Snapshot

Pop Using Linked List Node:
- Check if list is empty (head == null)
- Save data from head node
- Move head pointer to next node
- Delete old head node to free memory
- Return saved data
- Return -1 if list empty

Full Transcript

This concept shows how to remove the first node from a linked list safely. First, we check if the list is empty by seeing if the head pointer is null. If empty, we return -1. Otherwise, we save the data from the head node. Then, we move the head pointer to the next node in the list. After that, we delete the old head node to free memory. Finally, we return the saved data. This process updates the linked list by removing the first element and keeping the rest intact.

Practice

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

easy

A. Removes the first node and returns its value

B. Adds a new node at the end

C. Deletes all nodes in the list

D. Finds a node with a specific value

Solution

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

   Pop means removing the top or first element from the list, similar to popping from a stack.

2. Step 2: Identify what happens to the first node

   The first node is removed and its value is returned, and the head pointer moves to the next node.

3. Final Answer:

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

4. Quick Check:

   Pop = Remove first node [OK]

Hint: Pop removes the first node like taking the top card [OK]

Common Mistakes:

• Thinking pop adds a node
• Confusing pop with delete all
• Assuming pop searches for a value

2. Which of the following is the correct syntax to remove the first node in a singly linked list in C++?

easy

A. delete head->next;

B. head->next = head;

C. head = head->next;

D. head = NULL;

Solution

1. Step 1: Understand how to move head pointer

   To remove the first node, head must point to the second node, which is head->next.

2. Step 2: Check each option

   head = head->next; correctly assigns head to head->next. head->next = head; reverses pointers incorrectly. delete head->next; deletes the second node, not the first. head = NULL; sets head to NULL, deleting entire list.

3. Final Answer:

   head = head->next; -> Option C

4. Quick Check:

   Move head to next node = head = head->next; [OK]

Hint: Pop means head points to next node [OK]

Common Mistakes:

• Deleting wrong node
• Setting head to NULL incorrectly
• Reversing pointer assignment

3. What will be the output after executing the following code?

struct Node {
  int data;
  Node* next;
};

Node* head = new Node{10, nullptr};
head->next = new Node{20, nullptr};

int popValue = head->data;
head = head->next;

std::cout << popValue << " -> " << (head ? std::to_string(head->data) : "null") << std::endl;

medium

A. 20 -> 10

B. 10 -> 20

C. 10 -> null

D. 20 -> null

Solution

1. Step 1: Trace initial list nodes

   head points to node with data 10, which points to node with data 20.

2. Step 2: Execute pop operation

   popValue stores 10 (head's data), then head moves to next node (20).

3. Step 3: Print values

   Output prints popValue (10) and head's data (20), separated by ' -> '.

4. Final Answer:

   10 -> 20 -> Option B

5. Quick Check:

   Pop value = 10, new head = 20 [OK]

Hint: Pop returns old head data, head moves to next node [OK]

Common Mistakes:

• Printing new head data as pop value
• Confusing order of output
• Not updating head pointer

4. Identify the error in the following pop function for a singly linked list:

int pop(Node*& head) {
  int val = head->data;
  delete head;
  head = head->next;
  return val;
}

medium

A. Accessing head->next after deleting head

B. Not checking if head is NULL before pop

C. Returning value before deleting node

D. Function should return void

Solution

1. Step 1: Analyze order of operations

   The code deletes head node first, then tries to access head->next, which is invalid after deletion.

2. Step 2: Identify correct order

   Must save head->next before deleting head to avoid accessing freed memory.

3. Final Answer:

   Accessing head->next after deleting head -> Option A

4. Quick Check:

   Delete before accessing next = error [OK]

Hint: Save next pointer before deleting node [OK]

Common Mistakes:

• Deleting node before saving next
• Ignoring NULL head check
• Wrong return type assumption

5. Given a linked list with nodes containing values [5, 10, 15], write the correct sequence of operations to pop all nodes and print their values in order using C++ pointers.
Which code snippet correctly implements this?

hard

A. while(head) { int val = head->data; head = head->next; std::cout << val << " "; }

B. while(head) { std::cout << head->data << " "; delete head; }

C. while(head) { delete head; head = head->next; }

D. while(head) { int val = head->data; Node* temp = head; head = head->next; delete temp; std::cout << val << " "; }

Solution

1. Step 1: Understand safe pop and print

   We must print the current node's data, save the current node pointer, move head to next, then delete the saved node.

2. Step 2: Check each option

   while(head) { int val = head->data; head = head->next; std::cout << val << " "; } does not delete nodes, causing memory leak. while(head) { int val = head->data; Node* temp = head; head = head->next; delete temp; std::cout << val << " "; } correctly saves node in temp, moves head, deletes temp, and prints value. while(head) { delete head; head = head->next; } deletes head before moving, causing invalid access. while(head) { std::cout << head->data << " "; delete head; } deletes head but does not move head pointer, causing infinite loop or crash.

3. Final Answer:

   while(head) { int val = head->data; Node* temp = head; head = head->next; delete temp; std::cout << val << " "; } -> Option D

4. Quick Check:

   Save node, move head, delete node, print value [OK]

Hint: Save node before delete, move head after [OK]

Common Mistakes:

• Deleting node before moving head
• Not deleting nodes causing leaks
• Not moving head pointer