Reverse a Doubly Linked List in DSA Typescript - Build from Scratch

Reverse a Doubly Linked List

📖 Scenario: You are working on a music playlist app. The playlist is stored as a doubly linked list where each node holds a song name. Sometimes, users want to reverse the order of songs to listen backward.

🎯 Goal: Build a doubly linked list with given songs, then reverse the list by swapping the links, and finally print the reversed list.

📋 What You'll Learn

Create a doubly linked list with exact songs: 'Song1', 'Song2', 'Song3', 'Song4'

Create a variable called head pointing to the first node

Write a function called reverseDoublyLinkedList that reverses the list by swapping next and prev pointers

Print the reversed list starting from the new head using console.log

💡 Why This Matters

🌍 Real World

Doubly linked lists are used in music players, browsers (history), and undo-redo features where moving forward and backward is needed.

💼 Career

Understanding linked lists and how to reverse them is a common interview question and helps in mastering pointer manipulation and data structure fundamentals.

1

Create the doubly linked list nodes

Create a class called Node with properties data (string), next (Node or null), and prev (Node or null). Then create four nodes with data values 'Song1', 'Song2', 'Song3', and 'Song4'. Link them to form a doubly linked list where node1 is the head.

DSA Typescript

// Define Node class and create nodes for 'Song1', 'Song2', 'Song3', 'Song4'
// Link nodes to form doubly linked list
// Your code here

Hint

Remember to set both next and prev pointers for each node to link them properly.

2

Set the head of the list

Create a variable called head and assign it to node1, which is the first node of the doubly linked list.

DSA Typescript

class Node {
    data: string;
    next: Node | null;
    prev: Node | null;

    constructor(data: string) {
        this.data = data;
        this.next = null;
        this.prev = null;
    }
}

const node1 = new Node('Song1');
const node2 = new Node('Song2');
const node3 = new Node('Song3');
const node4 = new Node('Song4');

node1.next = node2;
node2.prev = node1;
node2.next = node3;
node3.prev = node2;
node3.next = node4;
node4.prev = node3;

// Create a variable called head and assign node1 to it
// Your code here

Hint

The head variable points to the first node in the list.

3

Write the function to reverse the doubly linked list

Write a function called reverseDoublyLinkedList that takes head of type Node | null and returns the new head after reversing. Inside the function, use a while loop to swap the next and prev pointers of each node. Update head to the last node processed and return it.

DSA Typescript

class Node {
    data: string;
    next: Node | null;
    prev: Node | null;

    constructor(data: string) {
        this.data = data;
        this.next = null;
        this.prev = null;
    }
}

const node1 = new Node('Song1');
const node2 = new Node('Song2');
const node3 = new Node('Song3');
const node4 = new Node('Song4');

node1.next = node2;
node2.prev = node1;
node2.next = node3;
node3.prev = node2;
node3.next = node4;
node4.prev = node3;

const head = node1;

// Write function reverseDoublyLinkedList(head: Node | null): Node | null
// Your code here

Hint

Swap next and prev for each node. Move to the original next node using the updated pointers.

4

Print the reversed doubly linked list

Call reverseDoublyLinkedList with head and store the result in reversedHead. Then use a while loop to print the data of each node starting from reversedHead, separated by ' -> '. End with 'null' using console.log.

DSA Typescript

class Node {
    data: string;
    next: Node | null;
    prev: Node | null;

    constructor(data: string) {
        this.data = data;
        this.next = null;
        this.prev = null;
    }
}

const node1 = new Node('Song1');
const node2 = new Node('Song2');
const node3 = new Node('Song3');
const node4 = new Node('Song4');

node1.next = node2;
node2.prev = node1;
node2.next = node3;
node3.prev = node2;
node3.next = node4;
node4.prev = node3;

const head = node1;

function reverseDoublyLinkedList(head: Node | null): Node | null {
    let current = head;
    let temp: Node | null = null;

    while (current !== null) {
        temp = current.prev;
        current.prev = current.next;
        current.next = temp;
        head = current;
        current = current.prev;
    }

    return head;
}

// Call reverseDoublyLinkedList and print the reversed list
// Your code here

Hint

Traverse from the new head and join each node's data with ' -> '. End with 'null'.

Practice

1. What is the main purpose of reversing a doubly linked list?

easy

A. To sort the nodes in ascending order

B. To reverse the order of nodes so the last becomes first and vice versa

C. To delete all nodes from the list

D. To convert it into a singly linked list

Solution

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

   A doubly linked list has nodes with pointers to both previous and next nodes.

2. Step 2: Purpose of reversing

   Reversing swaps the direction so the last node becomes the first and the first becomes the last.

3. Final Answer:

   To reverse the order of nodes so the last becomes first and vice versa -> Option B

4. Quick Check:

   Reversing = changing node order [OK]

Hint: Reversing flips the list order, not deleting or sorting [OK]

Common Mistakes:

• Confusing reversing with deleting nodes
• Thinking reversing sorts the list
• Assuming it converts to singly linked list

2. Which of the following TypeScript code snippets correctly swaps the next and prev pointers of a node in a doubly linked list?

easy

A. let temp = node.next; node.next = node.prev; node.prev = temp;

B. node.next = node.next; node.prev = node.prev;

C. node.prev = node.next; node.next = node.prev;

D. let temp = node.prev; node.prev = node.next; node.next = temp.next;

Solution

1. Step 1: Identify correct swapping logic

   Swapping pointers requires a temporary variable to hold one pointer before assignment.

2. Step 2: Check each option

   let temp = node.next; node.next = node.prev; node.prev = temp; correctly uses a temp variable to swap next and prev. Others either assign incorrectly or lose references.

3. Final Answer:

   let temp = node.next; node.next = node.prev; node.prev = temp; -> Option A

4. Quick Check:

   Swap pointers using temp variable [OK]

Hint: Use a temp variable to swap pointers safely [OK]

Common Mistakes:

• Assigning pointers without temp causes data loss
• Mixing up next and prev assignments
• Using incorrect pointer references

3. What will be the output of the following TypeScript code that reverses a doubly linked list?

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

function reverse(head: Node | null): Node | null {
  let current = head;
  let temp: Node | null = null;
  while (current !== null) {
    temp = current.prev;
    current.prev = current.next;
    current.next = temp;
    current = current.prev;
  }
  if (temp !== null) {
    head = temp.prev;
  }
  return head;
}

// Create list: 1 <-> 2 <-> 3
const n1 = new Node(1);
const n2 = new Node(2);
n1.next = n2; n2.prev = n1;
const n3 = new Node(3);
n2.next = n3; n3.prev = n2;

const newHead = reverse(n1);

// Print reversed list
let result = [];
let node = newHead;
while (node !== null) {
  result.push(node.data);
  node = node.next;
}
console.log(result);

medium

A. []

B. [1, 2, 3]

C. [3, 2, 1]

D. [3, 1, 2]

Solution

1. Step 1: Trace pointer swaps in the loop

   Each node swaps its next and prev. The loop moves forward using the old next pointer now in prev.

2. Step 2: Update head to new first node

   After loop, head is updated to last node processed, which is node with data 3.

3. Final Answer:

   [3, 2, 1] -> Option C

4. Quick Check:

   Reversed list order = [3, 2, 1] [OK]

Hint: Reversing swaps pointers and updates head to last node [OK]

Common Mistakes:

• Not updating head after reversal
• Confusing next and prev in loop
• Printing original list order

4. Identify the bug in this TypeScript function that attempts to reverse a doubly linked list:

function reverse(head: Node | null): Node | null {
  let current = head;
  while (current !== null) {
    let temp = current.next;
    current.next = current.prev;
    current.prev = temp;
    current = current.next;
  }
  return head;
}

medium

A. The function should return current instead of head

B. The pointers next and prev are swapped incorrectly

C. The loop condition should be current.next !== null

D. The function does not update the head to the new first node after reversal

Solution

1. Step 1: Check pointer swapping logic

   The swapping of next and prev is correct using a temp variable.

2. Step 2: Verify head update after reversal

   The function returns original head without updating it to the new first node, so the reversed list head is incorrect.

3. Final Answer:

   The function does not update the head to the new first node after reversal -> Option D

4. Quick Check:

   Head must update to new first node [OK]

Hint: Always update head to last processed node after reversal [OK]

Common Mistakes:

• Forgetting to update head after reversal
• Returning old head instead of new head
• Incorrect loop condition causing infinite loop

5. Given a doubly linked list with nodes containing integer data, how would you modify the reversal function to also count how many nodes were reversed and return both the new head and the count?

hard

A. Modify the function to use an object return type like { head: Node | null, count: number }, increment count inside the loop, and update head after reversal.

B. Add a global variable to count nodes and return only the new head.

C. Count nodes before reversal and return the count without changing the function.

D. Return an array with the reversed list nodes and count without modifying the reversal logic.

Solution

1. Step 1: Add a counter variable inside the reversal function

   Initialize a count variable to zero and increment it each time a node is processed in the reversal loop.

2. Step 2: Return both new head and count

   After reversal, return an object containing the updated head and the count of nodes reversed.

3. Final Answer:

   Modify the function to use an object return type like { head: Node | null, count: number }, increment count inside the loop, and update head after reversal. -> Option A

4. Quick Check:

   Return head and count together after reversal [OK]

Hint: Use an object return type to return multiple values [OK]

Common Mistakes:

• Using global variables instead of local counters
• Not updating head before returning
• Returning count without reversing