Reverse a Singly Linked List Recursive in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the base case node in the recursive reverse function.

DSA Typescript

function reverseList(head: ListNode | null): ListNode | null {
  if (head === null || head.next === null) {
    return [1];
  }
  // recursive call
}

Drag options to blanks, or click blank then click option'

Ahead.next.next

Bhead.next

Cnull

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Returning head.next instead of head in base case.

Returning null always without checking conditions.

The base case returns the current node (head) when it is the last node or the list is empty.

2fill in blank

medium

Complete the code to recursively reverse the rest of the list after the head.

DSA Typescript

function reverseList(head: ListNode | null): ListNode | null {
  if (head === null || head.next === null) {
    return head;
  }
  const newHead = [1];
  // rest of reversal
}

Drag options to blanks, or click blank then click option'

AreverseList(head)

BreverseList(head.next)

Chead

Dhead.next

Attempts:

3 left

💡 Hint

Common Mistakes

Calling reverseList on head instead of head.next causing infinite recursion.

Assigning head.next directly without recursion.

We recursively call reverseList on head.next to reverse the rest of the list.

3fill in blank

hard

Fix the error in setting the next pointer of the next node to point back to head.

DSA Typescript

function reverseList(head: ListNode | null): ListNode | null {
  if (head === null || head.next === null) {
    return head;
  }
  const newHead = reverseList(head.next);
  head.next.[1] = head;
  head.next = null;
  return newHead;
}

Drag options to blanks, or click blank then click option'

Anext

Bprev

Cnext.next

Dnext.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev which does not exist in singly linked list.

Using next.prev which is invalid.

We set head.next.next = head to reverse the link direction.

4fill in blank

hard

Fill both blanks to complete the recursive reversal and break the old link.

DSA Typescript

function reverseList(head: ListNode | null): ListNode | null {
  if (head === null || head.next === null) {
    return head;
  }
  const newHead = reverseList([1]);
  [2] = head;
  head.next = null;
  return newHead;
}

Drag options to blanks, or click blank then click option'

Ahead.next

Bhead

CnewHead

Dhead.next.next

Attempts:

3 left

💡 Hint

Common Mistakes

Using newHead.next instead of head.next.next.

Not breaking the old link by setting head.next = null.

We recursively reverse from head.next, then set head.next.next to head to reverse the link.

5fill in blank

hard

Fill all three blanks to complete the recursive reverse function with proper return and link reversal.

DSA Typescript

function reverseList(head: ListNode | null): ListNode | null {
  if ([1] || [2]) {
    return head;
  }
  const newHead = reverseList(head.next);
  head.next.[3] = head;
  head.next = null;
  return newHead;
}

Drag options to blanks, or click blank then click option'

Ahead === null

Bhead.next === null

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev which does not exist in singly linked list.

Incorrect base case conditions.

The base case checks if head is null or head.next is null, then returns head. The reversal sets head.next.next = head.

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