Delete Node at Beginning in DSA C++

Mental Model

Removing the first item means moving the start pointer to the next item and forgetting the old first one.

Analogy: Imagine a line of people holding hands. To remove the first person, you just let go of their hand and start holding the next person's hand instead.

head -> 1 -> 2 -> 3 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, delete node at beginning

Goal: Remove the first node so the list starts from the second node

Step 1: store current head (node with value 1) in temp

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

Why: We keep the first node temporarily to delete it safely

Step 2: move head pointer to next node (node with value 2)

head -> 2 -> 3 -> null, temp -> 1 -> null

Why: We update head to point to the new first node

Step 3: delete temp node (old first node)

head -> 2 -> 3 -> null

Why: We remove the old first node from memory

Result:

head -> 2 -> 3 -> null

Annotated Code

DSA C++

#include <iostream>
using namespace std;

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

class LinkedList {
public:
    Node* head;
    LinkedList() : head(nullptr) {}

    void insertAtEnd(int val) {
        Node* newNode = new Node(val);
        if (!head) {
            head = newNode;
            return;
        }
        Node* temp = head;
        while (temp->next) {
            temp = temp->next;
        }
        temp->next = newNode;
    }

    void deleteAtBeginning() {
        if (!head) return; // empty list check
        Node* temp = head; // store current head
        head = head->next; // move head to next node
        delete temp; // delete old head
    }

    void printList() {
        Node* temp = head;
        while (temp) {
            cout << temp->data;
            if (temp->next) cout << " -> ";
            temp = temp->next;
        }
        cout << " -> null" << endl;
    }
};

int main() {
    LinkedList list;
    list.insertAtEnd(1);
    list.insertAtEnd(2);
    list.insertAtEnd(3);

    list.printList();
    list.deleteAtBeginning();
    list.printList();

    return 0;
}

if (!head) return; // empty list check

stop if list is empty to avoid errors

Node* temp = head; // store current head

keep old first node to delete it safely

head = head->next; // move head to next node

update head pointer to new first node

delete temp; // delete old head

free memory of removed node

Output

1 -> 2 -> 3 -> null

2 -> 3 -> null

Complexity Analysis

Time: O(1) because we only change pointers at the start without traversing the list

Space: O(1) because no extra space is used except a temporary pointer

vs Alternative: Compared to deleting at the end which requires O(n) traversal, deleting at beginning is faster and simpler

Edge Cases

empty list

function returns immediately without error

DSA C++

if (!head) return; // empty list check

single node list

head becomes null after deletion, list becomes empty

DSA C++

head = head->next; // move head to next node

When to Use This Pattern

When you need to remove the first item quickly from a list, use the delete at beginning pattern because it only needs pointer updates without traversal.

Common Mistakes

Mistake: Not checking if the list is empty before deleting causes errors.
Fix: Add a check to return immediately if head is null.

Mistake: Not updating the head pointer to the next node before deleting causes loss of access to the rest of the list.
Fix: Always move head to head->next before deleting the old head.

Summary

Deletes the first node of a linked list by moving the head pointer to the next node and deleting the old first node.

Use when you want to remove the front element quickly without traversing the list.

The key is to update the head pointer before deleting the old node to keep the list connected.

Practice

1. What happens when you delete the first node of a singly linked list?

easy

A. The head pointer moves to the second node, and the first node is removed.

B. The last node is deleted instead of the first.

C. The list becomes empty regardless of its size.

D. The second node is deleted, but the head remains the same.

Solution

1. Step 1: Understand the head pointer role

   The head pointer points to the first node in the list.

2. Step 2: Delete the first node by moving head

   To delete the first node, update head to point to the second node, removing the first node from the list.

3. Final Answer:

   The head pointer moves to the second node, and the first node is removed. -> Option A

4. Quick Check:

   Deleting first node means head moves forward [OK]

Hint: Deleting first node means head points to next node [OK]

Common Mistakes:

• Deleting the wrong node (not the first)
• Not updating the head pointer
• Assuming list becomes empty always
• Confusing first and second nodes

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

easy

A. head->next = head; delete head;

B. head = head->next; delete head;

C. delete head; head = head->next;

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

Solution

1. Step 1: Store the current head node

   We save the current head in a temporary pointer to delete it safely later.

2. Step 2: Move head to next node and delete old head

   Update head to point to the next node, then delete the old head using the temporary pointer.

3. Final Answer:

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

4. Quick Check:

   Save old head, move head, delete old head [OK]

Hint: Save old head before deleting to avoid losing access [OK]

Common Mistakes:

• Deleting head before moving pointer
• Not saving old head before deleting
• Assigning head incorrectly
• Deleting head twice

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

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

Node* head = new Node{10, new Node{20, new Node{30, nullptr}}};

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

Node* curr = head;
while(curr != nullptr) {
  std::cout << curr->data << " -> ";
  curr = curr->next;
}
std::cout << "null";

medium

A. 20 -> 30 -> null

B. null

C. 30 -> null

D. 10 -> 20 -> 30 -> null

Solution

1. Step 1: Initial linked list setup

   The list starts as 10 -> 20 -> 30 -> null.

2. Step 2: Delete first node and update head

   We save head (10) in temp, move head to 20, then delete node 10.

3. Step 3: Print remaining list

   Starting from new head (20), we print 20 -> 30 -> null.

4. Final Answer:

   20 -> 30 -> null -> Option A

5. Quick Check:

   First node deleted, list starts at 20 [OK]

Hint: After deleting head, list starts from next node [OK]

Common Mistakes:

• Printing deleted node data
• Not updating head pointer
• Deleting wrong node
• Forgetting to print null at end

4. Identify the error in the following code snippet that deletes the first node of a linked list:

void deleteFirstNode(Node* head) {
  Node* temp = head;
  head = head->next;
  delete temp;
}

medium

A. head->next is not a valid pointer.

B. Deleting temp before moving head causes runtime error.

C. The function does not update the original head pointer outside the function.

D. No error; code works correctly.

Solution

1. Step 1: Understand pointer passing in C++

   Function receives head by value, so changes inside do not affect original head.

2. Step 2: Identify missing update to original head

   Updating local head pointer does not update the caller's head, so list head remains unchanged.

3. Final Answer:

   The function does not update the original head pointer outside the function. -> Option C

4. Quick Check:

   Pointer passed by value, original head unchanged [OK]

Hint: Pass head by reference to update original pointer [OK]

Common Mistakes:

• Assuming head changes outside function
• Deleting node before moving head
• Confusing pointer and value passing
• Ignoring nullptr checks

5. Given a singly linked list, write the correct approach to delete the first node safely and update the head pointer in C++. Which of the following code snippets correctly achieves this?

hard

A. void deleteFirstNode(Node* head) {
Node* temp = head;
head = head->next;
delete temp;
}

B. void deleteFirstNode(Node*& head) {
if (head == nullptr) return;
Node* temp = head;
head = head->next;
delete temp;
}

C. void deleteFirstNode(Node** head) {
Node* temp = *head;
*head = (*head)->next;
delete temp;
}

D. void deleteFirstNode(Node head) {
head = head.next;
delete head;
}

Solution

1. Step 1: Pass head by reference to modify original pointer

   Using Node*& head allows the function to update the caller's head pointer directly.

2. Step 2: Check for empty list and delete first node safely

   Check if head is nullptr to avoid errors, then move head to next node and delete old first node.

3. Final Answer:

   void deleteFirstNode(Node*& head) { if (head == nullptr) return; Node* temp = head; head = head->next; delete temp; } -> Option B

4. Quick Check:

   Reference parameter updates head safely [OK]

Hint: Use reference parameter to update head pointer [OK]

Common Mistakes:

• Passing head by value instead of reference
• Not checking for empty list
• Deleting node without updating head
• Using incorrect pointer syntax