Delete from End of Doubly Linked List in DSA C

Mental Model

To remove the last item, find the tail and unlink it from the list, updating the previous node to be the new tail.

Analogy: Imagine a train of connected cars. To remove the last car, you detach it from the second last car, making the second last car the new end of the train.

head -> 1 ↔ 2 ↔ 3 ↔ 4 ↔ 5 ← tail
null ← 1 ↔ 2 ↔ 3 ↔ 4 ↔ 5 -> null

Dry Run Walkthrough

Input: list: 1 ↔ 2 ↔ 3 ↔ 4 ↔ 5, delete from end

Goal: Remove the last node (5) and update the list tail

Step 1: start at head and move forward to find last node

1 ↔ 2 ↔ 3 ↔ 4 ↔ 5 -> null

Why: We need to find the last node to remove it

Step 2: last node (5) found, update previous node's next to null

1 ↔ 2 ↔ 3 ↔ 4 -> null  5 (removed)

Why: Disconnect last node from the list

Step 3: update tail pointer to previous node (4)

1 ↔ 2 ↔ 3 ↔ 4 ← tail -> null

Why: New tail is the node before the removed node

Result:

1 ↔ 2 ↔ 3 ↔ 4 -> null

Annotated Code

DSA C

#include <stdio.h>
#include <stdlib.h>

// Node structure for doubly linked list
typedef struct Node {
    int data;
    struct Node* prev;
    struct Node* next;
} Node;

// Function to create a new node
Node* createNode(int data) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = data;
    newNode->prev = NULL;
    newNode->next = NULL;
    return newNode;
}

// Function to append node at end
void append(Node** head_ref, int data) {
    Node* newNode = createNode(data);
    if (*head_ref == NULL) {
        *head_ref = newNode;
        return;
    }
    Node* temp = *head_ref;
    while (temp->next != NULL) {
        temp = temp->next;
    }
    temp->next = newNode;
    newNode->prev = temp;
}

// Function to delete node from end
void deleteFromEnd(Node** head_ref) {
    if (*head_ref == NULL) {
        return; // empty list
    }
    Node* temp = *head_ref;
    // move to last node
    while (temp->next != NULL) {
        temp = temp->next; // advance to tail
    }
    if (temp->prev != NULL) {
        temp->prev->next = NULL; // unlink last node
    } else {
        *head_ref = NULL; // list had one node, now empty
    }
    free(temp); // free memory
}

// Function to print list forward
void printList(Node* head) {
    Node* temp = head;
    while (temp != NULL) {
        printf("%d", temp->data);
        if (temp->next != NULL) {
            printf(" ↔ ");
        }
        temp = temp->next;
    }
    printf(" -> null\n");
}

int main() {
    Node* head = NULL;
    append(&head, 1);
    append(&head, 2);
    append(&head, 3);
    append(&head, 4);
    append(&head, 5);

    deleteFromEnd(&head);

    printList(head);
    return 0;
}

while (temp->next != NULL) { temp = temp->next; }

advance temp to last node to find tail

if (temp->prev != NULL) { temp->prev->next = NULL; } else { *head_ref = NULL; }

unlink last node and update tail or empty list

free(temp);

free memory of removed node

Output

1 ↔ 2 ↔ 3 ↔ 4 -> null

Complexity Analysis

Time: O(n) because we traverse the list from head to tail once to find the last node

Space: O(1) because we only use a few pointers and no extra data structures

vs Alternative: Compared to singly linked list, doubly linked list allows easier backward traversal but still requires O(n) to find tail if no tail pointer is maintained

Edge Cases

empty list

function returns immediately without changes

DSA C

if (*head_ref == NULL) { return; }

list with one node

node is removed and head is set to NULL

DSA C

else { *head_ref = NULL; }

When to Use This Pattern

When you need to remove the last element from a doubly linked list, look for a pattern that finds the tail and updates the previous node's next pointer to null.

Common Mistakes

Mistake: Not updating the previous node's next pointer to null, leaving a dangling pointer
Fix: Set temp->prev->next = NULL before freeing the last node

Mistake: Not handling the case when the list has only one node
Fix: Check if temp->prev is NULL and update head to NULL accordingly

Summary

Deletes the last node from a doubly linked list by unlinking it and updating pointers.

Use when you want to remove the tail element from a doubly linked list.

The key is to find the last node, unlink it from its previous node, and update the tail reference.

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