Reverse a Singly Linked List Recursive in DSA Typescript - Build from Scratch

Reverse a Singly Linked List Recursive

📖 Scenario: Imagine you have a chain of connected train cars, each linked to the next. You want to reverse the order of the cars so the last car becomes the first and the first becomes the last.

🎯 Goal: You will build a singly linked list and then reverse it using a recursive function. Finally, you will print the reversed list to see the new order.

📋 What You'll Learn

Create a singly linked list with exact nodes and values

Add a recursive function to reverse the linked list

Print the reversed linked list in the format: value -> value -> ... -> null

💡 Why This Matters

🌍 Real World

Reversing linked lists is useful in many applications like undo operations, navigation history, and data structure manipulation.

💼 Career

Understanding linked lists and recursion is important for software engineering interviews and working with low-level data structures.

1

Create the Singly Linked List

Create a class called ListNode with a constructor that takes value: number and next: ListNode | null. Then create three nodes with values 1, 2, and 3 linked together as head pointing to the first node.

DSA Typescript

// Define ListNode class and create linked list nodes
// Your code here

Hint

Start by defining the ListNode class with value and next. Then create nodes from last to first linking them.

2

Add Recursive Reverse Function

Create a function called reverseList that takes head: ListNode | null and returns ListNode | null. Use recursion to reverse the linked list.

DSA Typescript

class ListNode {
    value: number;
    next: ListNode | null;
    constructor(value: number, next: ListNode | null = null) {
        this.value = value;
        this.next = next;
    }
}

const node3 = new ListNode(3);
const node2 = new ListNode(2, node3);
const head = new ListNode(1, node2);

// Write the recursive function reverseList here
// Your code here

Hint

Use the base case when head is null or head.next is null. Then recursively reverse the rest and fix the links.

3

Reverse the Linked List

Call the reverseList function with head and store the result in a variable called reversedHead.

DSA Typescript

class ListNode {
    value: number;
    next: ListNode | null;
    constructor(value: number, next: ListNode | null = null) {
        this.value = value;
        this.next = next;
    }
}

const node3 = new ListNode(3);
const node2 = new ListNode(2, node3);
const head = new ListNode(1, node2);

function reverseList(head: ListNode | null): ListNode | null {
    if (head === null || head.next === null) {
        return head;
    }
    const newHead = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return newHead;
}

// Call reverseList with head and store in reversedHead
// Your code here

Hint

Simply call reverseList(head) and save the result in reversedHead.

4

Print the Reversed Linked List

Write a loop to print the values of the reversed linked list starting from reversedHead in the format: value -> value -> ... -> null.

DSA Typescript

class ListNode {
    value: number;
    next: ListNode | null;
    constructor(value: number, next: ListNode | null = null) {
        this.value = value;
        this.next = next;
    }
}

const node3 = new ListNode(3);
const node2 = new ListNode(2, node3);
const head = new ListNode(1, node2);

function reverseList(head: ListNode | null): ListNode | null {
    if (head === null || head.next === null) {
        return head;
    }
    const newHead = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return newHead;
}

const reversedHead = reverseList(head);

// Print the reversed linked list values
// Your code here

Hint

Use a while loop to go through each node starting from reversedHead. Add each value and an arrow to a string, then add null at the end and print it.

Practice

1. What is the main idea behind reversing a singly linked list using recursion?

easy

A. Create a new list and copy nodes in reverse order

B. Swap the values of the first and last nodes only

C. Reverse the rest of the list first, then fix the current node's next pointer

D. Iterate through the list and reverse pointers using a loop

Solution

1. Step 1: Understand recursion on linked list

   Recursion works by solving a smaller part first, here reversing the rest of the list beyond the current node.

2. Step 2: Fix current node's pointer after recursion

   After the rest is reversed, the current node's next node points back to it, reversing the link.

3. Final Answer:

   Reverse the rest of the list first, then fix the current node's next pointer -> Option C

4. Quick Check:

   Recursive reversal = reverse rest + fix pointer [OK]

Hint: Think: reverse smaller list first, then fix current link [OK]

Common Mistakes:

• Trying to swap node values instead of pointers
• Not fixing the current node's next pointer after recursion
• Using iteration instead of recursion

2. Which of the following is the correct base case for a recursive function that reverses a singly linked list?

easy

A. if (head === null) return head.next;

B. if (head.next === null) return null;

C. if (head === null) return head.next.next;

D. if (head === null || head.next === null) return head;

Solution

1. Step 1: Identify when recursion should stop

   The base case is when the list is empty or has only one node, so no reversal needed.

2. Step 2: Check the correct base case syntax

   Return the head if it is null or its next is null, meaning end or single node.

3. Final Answer:

   if (head === null || head.next === null) return head; -> Option D

4. Quick Check:

   Base case = empty or single node [OK]

Hint: Base case: empty or single node returns itself [OK]

Common Mistakes:

• Returning null when only one node exists
• Accessing head.next without checking if head is null
• Incorrectly returning head.next or head.next.next

3. What will be the output linked list after running this recursive reverse function on the list 1 -> 2 -> 3 -> null?

function reverseList(head: ListNode | null): ListNode | null {
  if (!head || !head.next) return head;
  const newHead = reverseList(head.next);
  head.next.next = head;
  head.next = null;
  return newHead;
}

medium

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. null

D. 2 -> 3 -> 1 -> null

Solution

1. Step 1: Trace recursion on list 1 -> 2 -> 3

   Recursion goes to node 3 (base case), then reverses links back: 3 points to 2, 2 points to 1.

2. Step 2: Update pointers to reverse list

   head.next.next = head sets 2's next to 1, head.next = null breaks old link, final reversed list is 3 -> 2 -> 1 -> null.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option A

4. Quick Check:

   Reversed list = 3 -> 2 -> 1 [OK]

Hint: Last node becomes new head after recursion [OK]

Common Mistakes:

• Not setting head.next to null causing cycles
• Returning original head instead of new head
• Confusing node values with pointers

4. Identify the bug in this recursive reverse function and choose the fix:

function reverseList(head: ListNode | null): ListNode | null {
  if (!head) return null;
  if (!head.next) return head;
  const newHead = reverseList(head.next);
  head.next = null;
  head.next.next = head;
  return newHead;
}

medium

A. Remove the base case for head.next === null

B. Swap the last two pointer assignments: set head.next.next before head.next = null

C. Return head instead of newHead

D. Add a loop to reverse pointers instead of recursion

Solution

1. Step 1: Analyze pointer assignments order

   Setting head.next = null before head.next.next = head breaks the chain because head.next becomes null.

2. Step 2: Correct order to fix pointers

   First set head.next.next = head to reverse link, then set head.next = null to break old link.

3. Final Answer:

   Swap the last two pointer assignments: set head.next.next before head.next = null -> Option B

4. Quick Check:

   Fix pointer order to avoid null errors [OK]

Hint: Set head.next.next before nullifying head.next [OK]

Common Mistakes:

• Nullifying head.next too early
• Ignoring base cases
• Returning wrong node

5. Given a singly linked list with nodes containing values [10, 20, 30, 40, 50], which recursive approach correctly reverses the list and returns the new head?

function reverseList(head: ListNode | null): ListNode | null {
  if (!head || !head.next) return head;
  const newHead = reverseList(head.next);
  head.next.next = head;
  head.next = null;
  return newHead;
}

What will be the final linked list after calling reverseList on the head node with value 10?

hard

A. 50 -> 40 -> 30 -> 20 -> 10 -> null

B. 10 -> 20 -> 30 -> 40 -> 50 -> null

C. null

D. 20 -> 30 -> 40 -> 50 -> 10 -> null

Solution

1. Step 1: Understand recursive reversal on list [10,20,30,40,50]

   The recursion reaches the last node (50), then reverses links step-by-step back to the first node.

2. Step 2: Trace pointer changes during recursion unwind

   Each node's next node points back to it, and its next pointer is set to null, resulting in reversed list 50 -> 40 -> 30 -> 20 -> 10 -> null.

3. Final Answer:

   50 -> 40 -> 30 -> 20 -> 10 -> null -> Option A

4. Quick Check:

   Recursive reverse flips list order [OK]

Hint: Last node becomes new head, links reversed back [OK]

Common Mistakes:

• Expecting original order after reversal
• Not setting head.next to null causing cycles
• Confusing node values with pointers