Insert at End of Doubly Linked List in DSA Javascript

Mental Model

A doubly linked list is a chain of nodes where each node knows the next and previous node. 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.

null ← 1 ↔ 2 ↔ 3 -> null
                 ↑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

null ← 1 ↔ 2 ↔ 3 -> null   4 (new node, not linked yet)

Why: We need a new node to insert at the end

Step 2: Set new node's previous pointer to current tail (3)

null ← 1 ↔ 2 ↔ 3 ←-> 4 -> null

Why: New node must know the node before it

Step 3: Set current tail's next pointer to new node

null ← 1 ↔ 2 ↔ 3 ↔ 4 -> null

Why: Current last node must point to new node

Step 4: Update tail pointer to new node (4)

null ← 1 ↔ 2 ↔ 3 ↔ 4 -> null
                     ↑tail

Why: Tail must point to the new last node

Result:

null ← 1 ↔ 2 ↔ 3 ↔ 4 -> null
                     ↑tail

Annotated Code

DSA Javascript

class Node {
  constructor(value) {
    this.value = value;
    this.next = null;
    this.prev = null;
  }
}

class DoublyLinkedList {
  constructor() {
    this.head = null;
    this.tail = null;
  }

  insertAtEnd(value) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      this.tail = newNode;
      return;
    }
    this.tail.next = newNode; // link current tail to new node
    newNode.prev = this.tail; // link new node back to current tail
    this.tail = newNode; // update tail to new node
  }

  printList() {
    let curr = this.head;
    let result = '';
    while (curr) {
      result += curr.value;
      if (curr.next) result += ' ↔ ';
      curr = curr.next;
    }
    console.log(result + ' -> null');
  }
}

// Driver code
const dll = new DoublyLinkedList();
dll.insertAtEnd(1);
dll.insertAtEnd(2);
dll.insertAtEnd(3);
dll.insertAtEnd(4);
dll.printList();

const newNode = new Node(value);

Create new node to insert

if (!this.head) { this.head = newNode; this.tail = newNode; return; }

Handle empty list by setting head and tail to new node

this.tail.next = newNode; // link current tail to new node

Link current tail's next to new node

newNode.prev = this.tail; // link new node back to current tail

Link new node's prev to current tail

this.tail = newNode; // update tail to new node

Update tail pointer to new node

Output

1 ↔ 2 ↔ 3 ↔ 4 -> null

Complexity Analysis

Time: O(1) because we directly access the tail and update pointers without traversal

Space: O(1) because we only create one new node regardless of list size

vs Alternative: Compared to singly linked list without tail pointer, this is faster since no traversal is needed to find the end

Edge Cases

Empty list

New node becomes both head and tail

DSA Javascript

if (!this.head) { this.head = newNode; this.tail = newNode; return; }

Single element list

New node links after the single node and tail updates

DSA Javascript

this.tail.next = newNode; // link current tail to new node

When to Use This Pattern

When you need to add an element at the end of a list and want fast insertion, use the insert at end pattern with a tail pointer in a doubly linked list.

Common Mistakes

Mistake: Forgetting to update the new node's prev pointer to the old tail
Fix: Always set newNode.prev = this.tail before updating tail

Mistake: Not updating the tail pointer to the new node after insertion
Fix: Set this.tail = newNode after linking nodes

Summary

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

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

The key is to link the new node to the current tail and then update the tail pointer.

Practice

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

easy

A. To reverse the order of the nodes in the list

B. To remove the first node from the list

C. To add a new element after the last node, updating both next and previous links

D. To find the middle node of the list

Solution

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

   A doubly linked list has nodes with pointers to both the next and previous nodes, 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 and updating the pointers accordingly.

3. Final Answer:

   To add a new element after the last node, updating both next and previous links -> Option C

4. Quick Check:

   Insert at end = Add after last node [OK]

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

Common Mistakes:

• Confusing insertion at end with deletion
• Not updating previous pointer of new node
• Trying to insert at beginning instead

2. Which of the following is the correct syntax to create a new node in JavaScript for a doubly linked list?

easy

A. const newNode = new Node(value);

B. const newNode = { value: data, nextNode: null };

C. const newNode = [value, null, null];

D. const newNode = { data: value, next: null, prev: null };

Solution

1. Step 1: Identify the node structure for doubly linked list

   A node should have data, next pointer, and previous pointer properties.

2. Step 2: Check each option for correct properties

   const newNode = { data: value, next: null, prev: null }; correctly uses data, next, and prev keys with null pointers initially.

3. Final Answer:

   const newNode = { data: value, next: null, prev: null }; -> Option D

4. Quick Check:

   Node = data + next + prev [OK]

Hint: Node needs data, next, and prev keys set [OK]

Common Mistakes:

• Using wrong property names like value instead of data
• Missing prev pointer in node
• Using array instead of object for node

3. Consider the following code snippet that inserts nodes at the end of a doubly linked list. What will be the printed list after inserting 10, 20, and 30?

class Node {
  constructor(data) {
    this.data = data;
    this.next = null;
    this.prev = null;
  }
}

class DoublyLinkedList {
  constructor() {
    this.head = null;
  }

  insertAtEnd(data) {
    const newNode = new Node(data);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let temp = this.head;
    while (temp.next) {
      temp = temp.next;
    }
    temp.next = newNode;
    newNode.prev = temp;
  }

  printList() {
    let temp = this.head;
    let result = '';
    while (temp) {
      result += temp.data + ' -> ';
      temp = temp.next;
    }
    result += 'null';
    console.log(result);
  }
}

const dll = new DoublyLinkedList();
dll.insertAtEnd(10);
dll.insertAtEnd(20);
dll.insertAtEnd(30);
dll.printList();

medium

A. null

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

C. 10 -> 20 -> null

D. 30 -> 20 -> 10 -> null

Solution

1. Step 1: Insert nodes one by one at the end

   First insert 10 as head, then 20 after 10, then 30 after 20.

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

   Traverse from head: 10 -> 20 -> 30 -> null is printed.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option B

4. Quick Check:

   Insert order = print order [OK]

Hint: Insert at end keeps order; print from head shows insertion order [OK]

Common Mistakes:

• Confusing insertion order with reversed output
• Not updating prev pointer causing broken links
• Printing from tail instead of head

4. The following code tries to insert a node at the end of a doubly linked list but causes an error. What is the mistake?

insertAtEnd(data) {
  const newNode = new Node(data);
  if (!this.head) {
    this.head = newNode;
  }
  let temp = this.head;
  while (temp.next) {
    temp = temp.next;
  }
  temp.next = newNode;
  newNode.prev = temp;
}

medium

A. Missing return after setting head when list is empty

B. Incorrectly assigning newNode.prev instead of newNode.next

C. Not creating newNode with correct constructor

D. Using while loop instead of for loop

Solution

1. Step 1: Analyze the empty list case

   If head is null, newNode is assigned as head but code continues to traverse and assign next pointers.

2. Step 2: Identify missing return causing duplicate insertion

   Without return after setting head, code tries to add newNode again after head, causing a cycle or error.

3. Final Answer:

   Missing return after setting head when list is empty -> Option A

4. Quick Check:

   Return after empty insert prevents duplicate links [OK]

Hint: Return immediately after empty list insertion [OK]

Common Mistakes:

• Forgetting to stop function after empty list insert
• Mixing up prev and next pointers
• Assuming loop type causes error

5. You want to insert a node with value 40 at the end of a doubly linked list that currently has nodes 10 -> 20 -> 30 -> null. After insertion, what should be the correct state of the list and the pointers of the new node?

hard

A. 10 -> 20 -> 30 -> 40 -> null, newNode.prev points to 30, newNode.next is null

B. 40 -> 10 -> 20 -> 30 -> null, newNode.prev is null, newNode.next points to 10

C. 10 -> 20 -> 30 -> null, newNode.prev and newNode.next both null

D. 10 -> 20 -> 30 -> 40 -> null, newNode.prev is null, newNode.next points to 30

Solution

1. Step 1: Insert new node at the end

   New node with value 40 is added after node 30, so 30's next points to new node.

2. Step 2: Set new node's pointers correctly

   New node's prev points to 30, and next is null because it's the last node.

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> null, newNode.prev points to 30, newNode.next is null -> Option A

4. Quick Check:

   Last node's next = null, newNode.prev = previous last node [OK]

Hint: New last node prev = old last node, next = null [OK]

Common Mistakes:

• Setting newNode.prev to null incorrectly
• Placing new node at the beginning instead
• Not updating previous last node's next pointer