Pop Using Linked List Node in DSA C++ - Build from Scratch

Pop Using Linked List Node

📖 Scenario: Imagine you have a stack of books represented as a linked list. You want to remove the top book from the stack.

🎯 Goal: Build a simple linked list with three nodes and then remove (pop) the first node, showing the list before and after popping.

📋 What You'll Learn

Create a linked list with three nodes containing values 10, 20, and 30

Create a pointer called head pointing to the first node

Write code to remove the first node (pop) from the linked list

Print the linked list before and after popping the first node

💡 Why This Matters

🌍 Real World

Linked lists are used in many programs to manage collections of items where you need to add or remove items efficiently from the front.

💼 Career

Understanding how to manipulate linked lists is important for software developers working with low-level data structures, memory management, and building efficient algorithms.

1

Create a linked list with three nodes

Create a struct called Node with an int data and a Node* next. Then create three nodes with values 10, 20, and 30. Link them so the first node points to the second, and the second points to the third. Create a pointer called head that points to the first node.

DSA C++

// Define Node struct and create three linked nodes
// Your code here

Hint

Remember to link nodes by setting the next pointer of one node to the next node.

2

Create a helper function to print the linked list

Add a function called printList that takes a Node* called head and prints the linked list values separated by -> and ending with null.

DSA C++

#include 

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

// Add printList function here
// Your code here

int main() {
    Node* node1 = new Node{10, nullptr};
    Node* node2 = new Node{20, nullptr};
    Node* node3 = new Node{30, nullptr};

    node1->next = node2;
    node2->next = node3;

    Node* head = node1;

    return 0;
}

Hint

Use a while loop to go through each node until nullptr.

3

Pop the first node from the linked list

In main, before returning, write code to remove the first node from the linked list by moving head to the next node and deleting the old first node.

DSA C++

#include 

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

void printList(Node* head) {
    Node* current = head;
    while (current != nullptr) {
        std::cout data  ";
        current = current->next;
    }
    std::cout next = node2;
    node2->next = node3;

    Node* head = node1;

    // Remove the first node (pop) here
    // Your code here

    return 0;
}

Hint

Save the current head in a temporary pointer, move head to the next node, then delete the old head.

4

Print the linked list before and after popping

In main, use printList(head) to print the linked list before popping the first node and after popping it.

DSA C++

#include 

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

void printList(Node* head) {
    Node* current = head;
    while (current != nullptr) {
        std::cout data  ";
        current = current->next;
    }
    std::cout next = node2;
    node2->next = node3;

    Node* head = node1;

    // Print list before popping
    // Your code here

    Node* temp = head;
    head = head->next;
    delete temp;

    // Print list after popping
    // Your code here

    return 0;
}

Hint

Call printList(head) before and after popping the first node to see the change.

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