Delete from End of Doubly Linked List in DSA C - Build from Scratch

Delete from End of Doubly Linked List

📖 Scenario: You are managing a playlist of songs using a doubly linked list. Each node contains a song number. You want to remove the last song from the playlist.

🎯 Goal: Build a doubly linked list with 3 nodes, then delete the last node, and finally print the updated list.

📋 What You'll Learn

Create a doubly linked list with exactly 3 nodes containing data 10, 20, and 30

Create a pointer called head pointing to the first node

Create a pointer called tail pointing to the last node

Write a function called deleteFromEnd that deletes the last node from the list

Print the list from head to tail after deletion

💡 Why This Matters

🌍 Real World

Doubly linked lists are used in music players, browsers, and undo-redo features where you need to move forward and backward easily.

💼 Career

Understanding linked list operations is fundamental for software developers working with data structures, memory management, and system programming.

1

Create the doubly linked list with 3 nodes

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

DSA C

// Define struct Node and create 3 linked nodes
// Your code here

Hint

Remember to link each node's next and prev pointers correctly.

2

Create the deleteFromEnd function

Create a function called deleteFromEnd that takes Node **head and Node **tail as parameters. It should delete the last node from the doubly linked list and update tail accordingly.

DSA C

#include 
#include 

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

void deleteFromEnd(Node **head, Node **tail) {
    // Your code here
}

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

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

    second->data = 20;
    second->prev = head;
    second->next = tail;

    tail->data = 30;
    tail->prev = second;
    tail->next = NULL;

    // Your code here

    return 0;
}

Hint

Check if the list is empty or has only one node before deleting.

3

Traverse and print the list from head to tail

Create a Node * pointer called current and set it to head. Use a while loop to traverse the list and print each node's data followed by ->. After the loop, print NULL.

DSA C

#include 
#include 

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

void deleteFromEnd(Node **head, Node **tail) {
    if (*tail == NULL) {
        return;
    }
    Node *temp = *tail;
    if ((*tail)->prev != NULL) {
        *tail = (*tail)->prev;
        (*tail)->next = NULL;
    } else {
        *head = NULL;
        *tail = NULL;
    }
    free(temp);
}

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

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

    second->data = 20;
    second->prev = head;
    second->next = tail;

    tail->data = 30;
    tail->prev = second;
    tail->next = NULL;

    deleteFromEnd(&head, &tail);

    Node *current = head;
    // Your code here

    return 0;
}

Hint

Use a while loop to move through the list until current is NULL.

4

Print the updated list after deletion

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

DSA C

#include 
#include 

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

void deleteFromEnd(Node **head, Node **tail) {
    if (*tail == NULL) {
        return;
    }
    Node *temp = *tail;
    if ((*tail)->prev != NULL) {
        *tail = (*tail)->prev;
        (*tail)->next = NULL;
    } else {
        *head = NULL;
        *tail = NULL;
    }
    free(temp);
}

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

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

    second->data = 20;
    second->prev = head;
    second->next = tail;

    tail->data = 30;
    tail->prev = second;
    tail->next = NULL;

    deleteFromEnd(&head, &tail);

    Node *current = head;
    while (current != NULL) {
        printf("%d -> ", current->data);
        current = current->next;
    }
    printf("NULL\n");

    // Your code here

    return 0;
}

Hint

Run the program to see the updated list after deleting the last node.

Practice

1. What happens to the tail pointer when you delete the last node from a doubly linked list?

easy

A. It becomes NULL immediately without checking.

B. It remains unchanged.

C. It points to the head node.

D. It moves to the previous node of the deleted node.

Solution

1. Step 1: Understand the tail pointer role

   The tail pointer always points to the last node in the doubly linked list.

2. Step 2: Update tail after deletion

   When the last node is deleted, the tail pointer must move to the previous node to keep pointing to the new last node.

3. Final Answer:

   It moves to the previous node of the deleted node. -> Option D

4. Quick Check:

   Tail points to previous node after deletion [OK]

Hint: Tail always points to last node; update to previous on delete [OK]

Common Mistakes:

• Setting tail to NULL without checking if list is empty
• Not updating tail pointer after deletion
• Pointing tail to head incorrectly

2. Which of the following is the correct syntax to delete the last node in a doubly linked list in C?

easy

A. tail = tail->prev; free(tail->next); tail->next = NULL;

B. free(tail); tail = tail->prev; tail->next = NULL;

C. tail->prev = tail; free(tail); tail = NULL;

D. tail->next = NULL; free(tail); tail = tail->prev;

Solution

1. Step 1: Free the last node correctly

   We first move tail to previous node, then free the old last node which is tail->next.

2. Step 2: Update tail's next pointer

   After freeing, set tail->next to NULL to mark new end of list.

3. Final Answer:

   tail = tail->prev; free(tail->next); tail->next = NULL; -> Option A

4. Quick Check:

   Move tail first, free old node, update next pointer [OK]

Hint: Move tail first, then free old node, update next pointer [OK]

Common Mistakes:

• Freeing tail before moving pointer
• Not setting tail->next to NULL
• Incorrect pointer assignments causing memory errors

3. What will be the output of the following code snippet after deleting the last node?

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

Node* tail = NULL;
// List: 10 <-> 20 <-> 30
// tail points to node with data 30

// Delete last node
Node* temp = tail;
tail = tail->prev;
free(temp);
tail->next = NULL;

// Print list from head
Node* head = tail;
while(head->prev) head = head->prev;
while(head) {
  printf("%d ", head->data);
  head = head->next;
}

medium

A. 10 20

B. 10 20 30

C. 20 30

D. 10 30

Solution

1. Step 1: Understand list before deletion

   Initial list is 10 <-> 20 <-> 30, tail points to 30.

2. Step 2: Delete last node and update pointers

   Last node 30 is freed, tail moves to 20, tail->next set to NULL.

3. Step 3: Print list from head

   Head is found by moving to start (10), printing nodes 10 and 20.

4. Final Answer:

   10 20 -> Option A

5. Quick Check:

   Last node removed, list prints 10 20 [OK]

Hint: After deletion, tail moves back; print from head to see updated list [OK]

Common Mistakes:

• Printing old tail node after deletion
• Not updating tail->next to NULL
• Starting print from tail instead of head

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

void deleteEnd(Node** tail) {
  if (*tail == NULL) return;
  Node* temp = *tail;
  *tail = (*tail)->prev;
  free(temp);
  (*tail)->next = NULL;
}

medium

A. Not freeing the node properly.

B. Dereferencing tail without checking if list becomes empty.

C. Not updating the head pointer.

D. Using double pointer for tail unnecessarily.

Solution

1. Step 1: Check for empty list after deletion

   If the list had only one node, after deletion tail becomes NULL, so (*tail)->next causes error.

2. Step 2: Fix by checking tail before dereference

   Must check if *tail is not NULL before setting (*tail)->next = NULL.

3. Final Answer:

   Dereferencing tail without checking if list becomes empty. -> Option B

4. Quick Check:

   Check tail NULL before accessing next pointer [OK]

Hint: Always check if tail is NULL before accessing its members [OK]

Common Mistakes:

• Dereferencing NULL pointer after deletion
• Ignoring empty list edge case
• Not freeing memory causing leaks

5. You have a doubly linked list with nodes: 5 <-> 15 <-> 25 <-> 35. After deleting the last node twice, what will be the printed list from head?

// Initial tail points to 35
// Delete last node twice
Node* temp;
temp = tail;
tail = tail->prev;
free(temp);
tail->next = NULL;

temp = tail;
tail = tail->prev;
free(temp);
tail->next = NULL;

// Print from head
Node* head = tail;
while(head->prev) head = head->prev;
while(head) {
  printf("%d ", head->data);
  head = head->next;
}

hard

A. 5 15 25

B. 15 25

C. 5 15

D. 25 35

Solution

1. Step 1: Understand initial list and tail

   List is 5 <-> 15 <-> 25 <-> 35, tail points to 35.

2. Step 2: Delete last node once

   Delete 35, tail moves to 25, tail->next = NULL.

3. Step 3: Delete last node second time

   Delete 25, tail moves to 15, tail->next = NULL.

4. Step 4: Print list from head

   Head found by moving to start (5), print nodes 5 and 15.

5. Final Answer:

   5 15 -> Option C

6. Quick Check:

   Two deletions remove last two nodes, list prints 5 15 [OK]

Hint: Each delete moves tail back; print from head to see current list [OK]

Common Mistakes:

• Not updating tail->next to NULL after each deletion
• Printing from tail instead of head
• Assuming deleted nodes still exist