Delete Node at End in DSA C - Build from Scratch

Delete Node at End in a Singly Linked List

📖 Scenario: Imagine you have a chain of boxes connected one after another. Each box holds a number and points to the next box. This is like a singly linked list. Sometimes, you want to remove the last box from the chain.

🎯 Goal: You will create a singly linked list with 3 nodes, then write code to remove the last node from the list, and finally print the list to see the result.

📋 What You'll Learn

Create a singly linked list with exactly 3 nodes containing values 10, 20, and 30 in that order

Create a pointer variable called head pointing to the first node

Write a function called deleteAtEnd that removes the last node from the list

Print the list after deletion showing the nodes connected with arrows like: 10 -> 20 -> NULL

💡 Why This Matters

🌍 Real World

Linked lists are used in many software systems to manage collections of items where size changes often, like music playlists or undo history.

💼 Career

Understanding how to manipulate linked lists is fundamental for software developers working with data structures, memory management, and system programming.

1

Create the singly linked list with 3 nodes

Create a struct called Node with an int data and a struct Node* next. Then create 3 nodes with values 10, 20, and 30. Link them so that head points to the first node.

DSA C

# Define struct Node and create 3 linked nodes with values 10, 20, 30
// Link nodes and set head pointer

Hint

Remember to use malloc to create nodes and link them by setting the next pointer.

2

Create the deleteAtEnd function

Write a function called deleteAtEnd that takes a pointer to Node* and removes the last node from the list. Handle the case when the list has only one node.

DSA C

#include 
#include 

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

void deleteAtEnd(Node** head) {
    // Your code here
}

int main() {
    Node* head = (Node*)malloc(sizeof(Node));
    Node* second = (Node*)malloc(sizeof(Node));
    Node* third = (Node*)malloc(sizeof(Node));

    head->data = 10;
    head->next = second;

    second->data = 20;
    second->next = third;

    third->data = 30;
    third->next = NULL;

    // Your code here

    return 0;
}

Hint

Use two pointers to find the second last node. Free the last node and set second last node's next to NULL.

3

Print the linked list

Write a function called printList that takes Node* head and prints the list nodes in the format: 10 -> 20 -> NULL. Then call printList(head) in main after deleting the last node.

DSA C

#include 
#include 

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

void deleteAtEnd(Node** head) {
    if (*head == NULL) {
        return;
    }
    if ((*head)->next == NULL) {
        free(*head);
        *head = NULL;
        return;
    }
    Node* temp = *head;
    while (temp->next->next != NULL) {
        temp = temp->next;
    }
    free(temp->next);
    temp->next = NULL;
}

void printList(Node* head) {
    // Your code here
}

int main() {
    Node* head = (Node*)malloc(sizeof(Node));
    Node* second = (Node*)malloc(sizeof(Node));
    Node* third = (Node*)malloc(sizeof(Node));

    head->data = 10;
    head->next = second;

    second->data = 20;
    second->next = third;

    third->data = 30;
    third->next = NULL;

    deleteAtEnd(&head);

    // Your code here

    return 0;
}

Hint

Use a loop to print each node's data followed by '->'. After the loop, print 'NULL'.

4

Print the final linked list after deletion

Run the program and observe the printed output after deleting the last node. The output should be: 10 -> 20 -> NULL

DSA C

# Your code here

Hint

Make sure you call printList(head) after deleting the last node to see the updated list.

Practice

1. What is the main purpose of the Delete Node at End operation in a singly linked list?

easy

A. To reverse the linked list

B. To add a new node at the end of the linked list

C. To delete the first node of the linked list

D. To remove the last node from the linked list

Solution

1. Step 1: Understand the operation's goal

   The operation aims to remove the last node in the linked list, which is the node that points to null.

2. Step 2: Identify the correct description

   Removing the last node means deleting the node at the end, not adding or reversing nodes.

3. Final Answer:

   To remove the last node from the linked list -> Option D

4. Quick Check:

   Delete Node at End = Remove last node [OK]

Hint: Deleting at end means removing the last node [OK]

Common Mistakes:

• Confusing delete at end with delete at beginning
• Thinking it adds a node instead of deleting
• Mixing up with reversing the list

2. Which of the following C code snippets correctly deletes the last node of a singly linked list named head?

easy

A. if (head == NULL || head->next == NULL) return; Node* temp = head; while (temp->next->next != NULL) temp = temp->next; free(temp->next); temp->next = NULL;

B. Node* temp = head; while (temp->next != NULL) temp = temp->next; free(temp);

C. free(head); head = NULL;

D. Node* temp = head; while (temp->next != NULL) temp = temp->next; temp->next = NULL;

Solution

1. Step 1: Check for empty or single-node list and traverse to second last node

   if (head == NULL || head->next == NULL) return; Node* temp = head; while (temp->next->next != NULL) temp = temp->next; free(temp->next); temp->next = NULL; checks if list is empty or single node, then moves to the node before last using temp->next->next != NULL.

2. Step 2: Free last node and update pointer

   It frees the last node and sets temp->next to NULL, correctly deleting the last node.

3. Final Answer:

   if (head == NULL || head->next == NULL) return; Node* temp = head; while (temp->next->next != NULL) temp = temp->next; free(temp->next); temp->next = NULL; -> Option A

4. Quick Check:

   Traverse to second last, free last, set next NULL = if (head == NULL || head->next == NULL) return; Node* temp = head; while (temp->next->next != NULL) temp = temp->next; free(temp->next); temp->next = NULL; [OK]

Hint: Stop at second last node before deleting last [OK]

Common Mistakes:

• Freeing node without updating previous node's next pointer
• Not checking for empty list
• Freeing head without traversal

3. What will be the output linked list after deleting the last node from the list: 10 -> 20 -> 30 -> 40 -> NULL using the standard delete at end operation?

medium

A. 10 -> 20 -> 30 -> 40 -> NULL

B. 10 -> 20 -> 30 -> NULL

C. 20 -> 30 -> 40 -> NULL

D. 10 -> 20 -> NULL

Solution

1. Step 1: Identify the last node

   The last node is 40, which points to NULL.

2. Step 2: Remove the last node and update the second last node's next pointer

   After deletion, 30 becomes the last node and its next pointer is set to NULL.

3. Final Answer:

   10 -> 20 -> 30 -> NULL -> Option B

4. Quick Check:

   Last node 40 removed = 10 -> 20 -> 30 -> NULL [OK]

Hint: Last node removed, second last points to NULL [OK]

Common Mistakes:

• Not updating second last node's next pointer
• Deleting wrong node (like first node)
• Output list unchanged

4. Consider the following code snippet to delete the last node of a singly linked list. What is the error?

void deleteEnd(Node** head) {
  if (*head == NULL) return;
  Node* temp = *head;
  while (temp->next != NULL) {
    temp = temp->next;
  }
  free(temp);
  temp = NULL;
}

medium

A. It does not update the second last node's next pointer to NULL

B. It frees the head node instead of the last node

C. It causes a segmentation fault by accessing NULL pointer

D. It deletes the first node instead of the last

Solution

1. Step 1: Analyze traversal logic

   The loop moves temp to the last node (where temp->next is NULL).

2. Step 2: Check pointer update after freeing

   It frees the last node but does not update the second last node's next pointer to NULL, leaving a dangling pointer.

3. Final Answer:

   It does not update the second last node's next pointer to NULL -> Option A

4. Quick Check:

   Missing next pointer update causes dangling pointer [OK]

Hint: Always update second last node's next to NULL after free [OK]

Common Mistakes:

• Freeing node without updating previous node's next
• Not handling empty list
• Incorrect loop condition causing crash

5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> 35 -> 45 -> NULL. Write the steps to delete the last node and then insert a new node with value 50 at the end. What will be the final list?

hard

A. 5 -> 15 -> 25 -> 35 -> NULL

B. 5 -> 15 -> 25 -> 35 -> 45 -> 50 -> NULL

C. 5 -> 15 -> 25 -> 35 -> 50 -> NULL

D. 50 -> 5 -> 15 -> 25 -> 35 -> NULL

Solution

1. Step 1: Delete the last node (45)

   Traverse to second last node (35), free last node (45), set 35's next to NULL.

2. Step 2: Insert new node with value 50 at end

   Traverse to last node (35), create new node with 50, set 35's next to new node, new node's next to NULL.

3. Final Answer:

   5 -> 15 -> 25 -> 35 -> 50 -> NULL -> Option C

4. Quick Check:

   Delete last, then add 50 at end = 5 -> 15 -> 25 -> 35 -> 50 -> NULL [OK]

Hint: Delete last node, then add new node at end [OK]

Common Mistakes:

• Not updating pointers correctly after deletion
• Appending new node without deleting last
• Inserting new node at wrong position