Delete from End of Doubly Linked List in DSA Python - Execution Trace

Concept Flow - Delete from End of Doubly Linked List

Start at tail node

↓

Check if list is empty?

| No↓

If only one node

| Yes↓

Set head and tail to None

No↓

Move tail pointer to previous node

↓

Set new tail's next to None

↓

Delete old tail node

↓

Done

Start from the tail node, check if list is empty or has one node, then update tail pointer and remove last node.

Execution Sample

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.prev = None
        self.next = None

class DoublyLinkedList:
    def __init__(self):
        self.head = None
        self.tail = None

    def delete_end(self):
        if not self.tail:
            return
        if self.head == self.tail:
            self.head = None
            self.tail = None
            return
        self.tail = self.tail.prev
        self.tail.next = None

Deletes the last node from a doubly linked list, updating pointers accordingly.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Start with list: 10 <-> 20 <-> 30 <-> 40	10 <-> 20 <-> 30 <-> 40	head → 10, tail → 40	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ data:40│ │ prev:∅ │←── │ prev:10│←── │ prev:20│←── │ prev:30│ │ next:20│──→ │ next:30│──→ │ next:40│──→ │ next:∅ │ └────────┘ └────────┘ └────────┘ └────────┘ head tail
2	Check if list is empty	10 <-> 20 <-> 30 <-> 40	No change	Same as step 1
3	Check if only one node	10 <-> 20 <-> 30 <-> 40	No change	Same as step 1
4	Move tail pointer to previous node (30)	10 <-> 20 <-> 30 <-> 40	tail: 40 → 30	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ data:40│ │ prev:∅ │←── │ prev:10│←── │ prev:20│←── │ prev:30│ │ next:20│──→ │ next:30│──→ │ next:40│──→ │ next:∅ │ └────────┘ └────────┘ └────────┘ └────────┘ head tail (moved here)
5	Set new tail's next to None	10 <-> 20 <-> 30	30.next: 40 → None	┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ prev:∅ │←── │ prev:10│←── │ prev:20│ │ next:20│──→ │ next:30│──→ │ next:∅ │ └────────┘ └────────┘ └────────┘ head tail
6	Old tail (40) is now disconnected and deleted	10 <-> 20 <-> 30	40 node removed	┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ prev:∅ │←── │ prev:10│←── │ prev:20│ │ next:20│──→ │ next:30│──→ │ next:∅ │ └────────┘ └────────┘ └────────┘ head tail
7	End of delete operation	10 <-> 20 <-> 30	Pointers stable	Same as step 6

💡 Tail moved to previous node and old tail disconnected, list updated

Variable Tracker

Variable	Start	After Step 4	After Step 5	Final
head	Node(10)	Node(10)	Node(10)	Node(10)
tail	Node(40)	Node(30)	Node(30)	Node(30)
tail.next	None	Node(40)	None	None
list size	4 nodes	4 nodes	3 nodes	3 nodes

Key Moments - 3 Insights

Why do we check if the list has only one node before deleting?

Because if there is only one node, deleting it means the list becomes empty. We must set both head and tail to None to avoid dangling pointers. See step 3 in execution_table.

Why do we set the new tail's next pointer to None?

Because the new tail is the last node now, so its next pointer must not point to any node. This disconnects the old tail. See step 5 in execution_table.

What happens to the old tail node after deletion?

It becomes disconnected from the list and is removed from memory (garbage collected). It no longer has any references. See step 6 in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, which node does the tail pointer move to?

A Node with data 20

B Node with data 30

C Node with data 40

D Node with data 10

💡 Hint

Check the 'Pointer Changes' column in step 4 of execution_table

At which step does the list size reduce from 4 to 3 nodes?

A Step 3

B Step 6

C Step 5

D Step 7

💡 Hint

Look at the 'Nodes in List' and 'Variable Tracker' after step 5

If the list had only one node, what would happen to head and tail after deletion?

A Both head and tail become None

B Head points to None, tail remains the node

C Both remain pointing to the single node

D Tail points to None, head remains the node

💡 Hint

Refer to step 3 and the key moment about single node deletion

Concept Snapshot

Delete from End of Doubly Linked List:
- Start at tail node
- If list empty: do nothing
- If one node: set head and tail to None
- Else move tail to tail.prev
- Set new tail.next to None
- Old tail node removed
Pointers updated to keep list consistent

Full Transcript

To delete the last node from a doubly linked list, we start at the tail. If the list is empty, we do nothing. If there is only one node, we set both head and tail to None, making the list empty. Otherwise, we move the tail pointer to the previous node and set the new tail's next pointer to None to disconnect the old tail. The old tail node is then removed from the list. This process updates pointers carefully to maintain the list's structure.