Pop Using Linked List Node in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'pop' operation do in a linked list?

It removes the first node (head) from the linked list and returns its value, effectively shortening the list by one node.

beginner

In a singly linked list, what pointer is updated when popping the head node?

The head pointer is updated to point to the next node after the current head.

intermediate

Why must we be careful about memory when popping a node in C++ linked list?

Because the popped node is removed from the list, we must delete it to free memory and avoid memory leaks.

beginner

What happens if you pop from an empty linked list?

Since there is no node to remove, the operation should handle this gracefully, often by returning a special value or indicating the list is empty.

intermediate

Show the basic steps to pop a node from a singly linked list in C++.

1. Check if head is null (empty list). 2. Save pointer to current head. 3. Move head to next node. 4. Store value from old head. 5. Delete old head node. 6. Return stored value.

What pointer does the head point to after popping the first node in a singly linked list?

A The next node after the popped node

B The previous node before the popped node

C It remains the same

D Null always

After popping, the head pointer moves to the next node in the list.

What must you do with the popped node in C++ to avoid memory leaks?

A Delete the node using delete operator

B Set the node pointer to null

C Do nothing, it frees automatically

D Copy the node to another variable

In C++, you must explicitly delete the popped node to free memory.

What is the return value of pop operation if the linked list is empty?

A Zero always

B The last node's value

C A special value or indication of empty list

D The head pointer

Pop on empty list should return a special value or indicate the list is empty.

Which node is removed during a pop operation on a singly linked list?

A The tail node

B The middle node

C A random node

D The head node

Pop removes the head node from the list.

What happens to the linked list size after a pop operation?

A It doubles

B It decreases by one

C It stays the same

D It increases by one

Popping removes one node, so the list size decreases by one.

Explain step-by-step how to pop a node from a singly linked list and why each step is important.

Think about what happens to the head pointer and memory.

Describe what happens internally in the linked list when you pop the first node.

Focus on pointer changes and memory.

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