Insert at End of Doubly Linked List in DSA Python

Mental Model

A doubly linked list has nodes connected forward and backward. To add a new node at the end, we link it after the last node and update pointers.

Analogy: Imagine a train where each carriage is connected to the one before and after. Adding a new carriage at the end means hooking it to the last carriage and making sure both know about each other.

head -> [1] ↔ [2] ↔ [3] -> null
null ← [1] ↔ [2] ↔ [3] ← tail

Dry Run Walkthrough

Input: list: 1 ↔ 2 ↔ 3, insert value 4 at end

Goal: Add a new node with value 4 at the end of the doubly linked list

Step 1: Create new node with value 4

head -> [1] ↔ [2] ↔ [3] -> null, new_node=[4] (isolated)

Why: We need a new node ready to insert

Step 2: Traverse from head to last node (3)

head -> [1] ↔ [2] ↔ [3 ↑] -> null, new_node=[4]

Why: We must find the last node to link new node after it

Step 3: Set last node's next pointer to new node

head -> [1] ↔ [2] ↔ [3] -> [4], new_node=[4]

Why: Link last node forward to new node

Step 4: Set new node's previous pointer to last node

head -> [1] ↔ [2] ↔ [3] ↔ [4] -> null

Why: Link new node backward to last node

Result:

head -> [1] ↔ [2] ↔ [3] ↔ [4] -> null

Annotated Code

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

    def insert_at_end(self, data):
        new_node = Node(data)  # create new node
        if self.head is None:  # empty list case
            self.head = new_node
            return
        curr = self.head
        while curr.next:  # traverse to last node
            curr = curr.next
        curr.next = new_node  # link last node to new node
        new_node.prev = curr  # link new node back to last node

    def __str__(self):
        nodes = []
        curr = self.head
        while curr:
            nodes.append(str(curr.data))
            curr = curr.next
        return ' ↔ '.join(nodes) + ' -> null'

# Driver code
dll = DoublyLinkedList()
dll.insert_at_end(1)
dll.insert_at_end(2)
dll.insert_at_end(3)
dll.insert_at_end(4)
print(dll)

new_node = Node(data) # create new node

create the new node to insert

if self.head is None: # empty list case

handle empty list by setting head to new node

while curr.next: # traverse to last node

move curr forward until last node

curr.next = new_node # link last node to new node

connect last node's next to new node

new_node.prev = curr # link new node back to last node

connect new node's prev to last node

Output

1 ↔ 2 ↔ 3 ↔ 4 -> null

Complexity Analysis

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

Space: O(1) because we only create one new node and use a few pointers

vs Alternative: Compared to inserting at the beginning (O(1)), inserting at end requires traversal, so it is slower for long lists

Edge Cases

empty list

new node becomes the head of the list

DSA Python

if self.head is None:  # empty list case

single element list

new node is linked after the only node, both nodes point to each other correctly

DSA Python

while curr.next:  # traverse to last node

When to Use This Pattern

When you need to add an element at the end of a doubly linked list, use the insert at end pattern to update both next and prev pointers correctly.

Common Mistakes

Mistake: Forgetting to update the new node's prev pointer to the last node
Fix: Always set new_node.prev = curr after linking curr.next = new_node

Mistake: Not handling the empty list case, causing errors when head is None
Fix: Check if head is None and set head to new node before traversal

Summary

Adds a new node at the end of a doubly linked list by updating next and prev pointers.

Use when you want to append data to the end of a doubly linked list.

Remember to link both directions: last node's next to new node and new node's prev to last node.

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