Insert at End of Doubly Linked List in DSA Python - Execution Trace

Concept Flow - Insert at End of Doubly Linked List

Create new node with data

↓

Is list empty? head == None

↓

Set head = new

↓

Traverse to last node

↓

Set last.next = new

↓

Set new.prev = last

↓

Done

Create a new node, check if list is empty, if yes set head to new node. Otherwise, traverse to last node, link last node's next to new node, and new node's prev to last node.

Execution Sample

DSA Python

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

def insert_end(head, data):
    new_node = Node(data)
    if head is None:
        head = new_node
        return head
    current = head
    while current.next:
        current = current.next
    current.next = new_node
    new_node.prev = current
    return head

# Example usage:
head = Node(10)
second = Node(20)
third = Node(30)
head.next = second
second.prev = head
second.next = third
third.prev = second

# Insert 40 at end
head = insert_end(head, 40)

Insert a new node with value 40 at the end of a doubly linked list starting at head.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Create new node with data=40	10 <-> 20 <-> 30	new_node.prev = None, new_node.next = None	┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ prev:∅ │ │ prev───┼──→ │ prev───┼──→ │ │ next───┼──→ │ next───┼──→ │ next:∅ │ └────────┘ └────────┘ └────────┘ head
2	Check if head is None	10 <-> 20 <-> 30	head is not None, proceed to traverse	Same as above
3	Traverse to last node (30)	10 <-> 20 <-> 30	current moves from 10 to 20 to 30	Same as above, pointer current at node 30
4	Set last.next = new_node	10 <-> 20 <-> 30	30.next = new_node	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ data:40│ │ prev:∅ │ │ prev───┼──→ │ prev───┼──→ │ prev:∅ │ │ next───┼──→ │ next───┼──→ │ next───┼──→ │ next:∅ │ └────────┘ └────────┘ └────────┘ └────────┘ head
5	Set new_node.prev = last node (30)	10 <-> 20 <-> 30 <-> 40	new_node.prev = 30	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ data:40│ │ prev:∅ │ │ prev───┼──→ │ prev───┼──→ │ prev───┼──→ 30 │ next───┼──→ │ next───┼──→ │ next───┼──→ │ next:∅ │ └────────┘ └────────┘ └────────┘ └────────┘ head
6	Insertion complete	10 <-> 20 <-> 30 <-> 40	Pointers stable	┌────────┐ ┌────────┐ ┌────────┐ ┌────────┐ │ data:10│ │ data:20│ │ data:30│ │ data:40│ │ prev:∅ │ │ prev───┼──→ │ prev───┼──→ │ prev───┼──→ │ next───┼──→ │ next───┼──→ │ next───┼──→ │ next:∅ │ └────────┘ └────────┘ └────────┘ └────────┘ head

💡 Insertion done after last node, new node's next is None

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 4	After Step 5	Final
head	Node(10)	Node(10)	Node(10)	Node(10)	Node(10)	Node(10)
current	None	None	Node(30)	Node(30)	Node(30)	Node(30)
new_node	None	Node(40)	Node(40)	Node(40)	Node(40)	Node(40)

Key Moments - 3 Insights

Why do we check if head is None before inserting?

If head is None, the list is empty, so the new node becomes the head. This is shown in step 2 of the execution_table where we decide to traverse or set head.

Why do we set new_node.prev to last node?

Because in a doubly linked list, each node points back to its previous node. Step 5 shows setting new_node.prev to the last node to maintain backward links.

How do we know where the last node is?

We traverse starting from head until current.next is None, meaning current is the last node. Step 3 shows this traversal.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'current' after step 3?

A Node with data 20

B Node with data 30

C Node with data 40

D None

💡 Hint

Check the 'current' variable in variable_tracker after step 3

At which step is the new node's prev pointer set to the last node?

A Step 5

B Step 4

C Step 2

D Step 6

💡 Hint

Look at the Pointer Changes column in execution_table rows

If the list was empty initially, what would be the visual state after insertion?

A No nodes, head is None

B Two nodes linked together

C Single node with prev and next as None, head points to it

D New node with next pointing to head

💡 Hint

Refer to concept_flow branch when head is None

Concept Snapshot

Insert at End of Doubly Linked List:
- Create new node with data
- If list empty (head is None), set head = new node
- Else traverse to last node
- Set last.next = new node
- Set new_node.prev = last node
- New node's next = None
- List size increases by 1

Full Transcript

To insert a node at the end of a doubly linked list, first create the new node with the given data. Check if the list is empty by seeing if head is None. If empty, set head to the new node. Otherwise, start at head and move through the list until reaching the last node (where next is None). Then set the last node's next pointer to the new node, and set the new node's prev pointer back to the last node. The new node's next pointer is None, marking it as the new tail. This maintains the doubly linked structure with forward and backward links.

Practice

1. What is the main purpose of inserting a node at the end of a doubly linked list?

easy

A. To reverse the list

B. To add a new element after the last node

C. To remove the first node

D. To find the middle node

Solution

1. Step 1: Understand the structure of a doubly linked list

   A doubly linked list has nodes connected forward and backward, allowing traversal in both directions.

2. Step 2: Purpose of insertion at the end

   Inserting at the end means adding a new node after the current last node, updating pointers accordingly.

3. Final Answer:

   To add a new element after the last node -> Option B

4. Quick Check:

   Insert at end = Add after last node [OK]

Hint: Insert at end means add after last node [OK]

Common Mistakes:

• Confusing insertion at end with deletion
• Thinking insertion reverses the list
• Assuming insertion happens at the start

2. Which of the following is the correct way to update the pointers when inserting a new node at the end of a doubly linked list in Python?

easy

A. new_node.prev = null; new_node.next = head; head.prev = new_node; head = new_node

B. new_node.next = head; new_node.prev = null; head.prev = new_node; head = new_node

C. new_node.next = null; new_node.prev = tail; tail.next = new_node; tail = new_node

D. new_node.prev = tail; new_node.next = tail; tail.next = new_node; tail = new_node

Solution

1. Step 1: Identify pointer updates for insertion at end

   The new node's next pointer should be null because it will be the last node. Its prev pointer should point to the current tail.

2. Step 2: Update the tail's next pointer and move tail

   The current tail's next pointer should point to the new node, then update tail to the new node.

3. Final Answer:

   new_node.next = null; new_node.prev = tail; tail.next = new_node; tail = new_node -> Option C

4. Quick Check:

   New node next = null, prev = tail [OK]

Hint: New node next is null, prev points to tail [OK]

Common Mistakes:

• Setting new_node.next to head instead of null
• Not updating tail pointer after insertion
• Confusing prev and next pointers

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

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 insert_end(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = self.tail = new_node
        else:
            self.tail.next = new_node
            new_node.prev = self.tail
            self.tail = new_node

    def print_list(self):
        current = self.head
        result = []
        while current:
            result.append(current.data)
            current = current.next
        print(' -> '.join(map(str, result)) + ' -> null')

dll = DoublyLinkedList()
dll.insert_end(10)
dll.insert_end(20)
dll.insert_end(30)
dll.print_list()

medium

A. 10 -> 20 -> 30 -> null

B. 30 -> 20 -> 10 -> null

C. 10 -> 20 -> null

D. null

Solution

1. Step 1: Insert nodes 10, 20, 30 at the end

   Each insert_end call adds a new node at the tail, linking nodes in order: 10, then 20, then 30.

2. Step 2: Print the list from head to tail

   print_list traverses from head, collecting data: 10, 20, 30, then prints with arrows and 'null' at end.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option A

4. Quick Check:

   Insert end adds at tail, print from head [OK]

Hint: Insert end adds at tail; print from head shows order [OK]

Common Mistakes:

• Confusing print order as reversed
• Missing last node in output
• Printing null instead of 'null'

4. Identify the error in the following code snippet that inserts a node at the end of a doubly linked list:

def insert_end(self, data):
    new_node = Node(data)
    if self.head is None:
        self.head = new_node
    else:
        self.tail.next = new_node
        new_node.prev = self.tail
        self.tail = new_node

medium

A. new_node.next should point to head

B. new_node.prev should be null when list is empty

C. self.head should be updated after insertion at end

D. Missing update of tail when list is empty

Solution

1. Step 1: Check tail update when list is empty

   When the list is empty, only head is set to new_node, but tail is not updated, so tail remains null.

2. Step 2: Correct tail assignment

   Tail must also be set to new_node when list is empty to keep tail pointer valid.

3. Final Answer:

   Missing update of tail when list is empty -> Option D

4. Quick Check:

   Empty list needs tail = new_node too [OK]

Hint: Update tail when list is empty [OK]

Common Mistakes:

• Not updating tail on empty list
• Setting new_node.next incorrectly
• Confusing head and tail updates

5. Given a doubly linked list with nodes containing values [5, 15, 25], you want to insert a new node with value 35 at the end. After insertion, what will be the correct sequence of node values when traversing from head to tail and tail to head respectively?

hard

A. 5 -> 15 -> 25 -> 35 -> null and 35 -> 25 -> 15 -> 5 -> null

B. 35 -> 5 -> 15 -> 25 -> null and 25 -> 15 -> 5 -> 35 -> null

C. 5 -> 15 -> 25 -> null and 35 -> null

D. 5 -> 15 -> 25 -> 35 -> null and 5 -> 15 -> 25 -> 35 -> null

Solution

1. Step 1: Insert 35 at the end of the list

   The new node with value 35 is added after the current tail (25), updating pointers accordingly.

2. Step 2: Traverse from head to tail and tail to head

   From head: 5 -> 15 -> 25 -> 35 -> null; From tail: 35 -> 25 -> 15 -> 5 -> null.

3. Final Answer:

   5 -> 15 -> 25 -> 35 -> null and 35 -> 25 -> 15 -> 5 -> null -> Option A

4. Quick Check:

   Forward and backward traversal show correct order [OK]

Hint: Forward order adds new node at end; backward reverses [OK]

Common Mistakes:

• Mixing forward and backward traversal order
• Forgetting to update prev pointer of new node
• Assuming new node is added at head