Reverse a Singly Linked List Recursive in DSA Typescript - Time & Space Complexity

Time Complexity: Reverse a Singly Linked List Recursive

O(n)

Understanding Time Complexity

We want to understand how the time needed to reverse a linked list grows as the list gets longer.

Specifically, how does the recursive method handle each node as the list size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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;
}
    

This code reverses a singly linked list by calling itself on the next node until it reaches the end, then rewires the links backward.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive call on each node of the list.
• How many times: Once per node, so exactly n times for a list of size n.

How Execution Grows With Input

Each node causes one recursive call, so the total steps grow directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 recursive calls and rewiring steps
100	About 100 recursive calls and rewiring steps
1000	About 1000 recursive calls and rewiring steps

Pattern observation: The work grows linearly as the list size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows directly in proportion to the number of nodes.

Common Mistake

[X] Wrong: "Recursive reversal is faster because it uses fewer lines of code."

[OK] Correct: The number of steps depends on the list size, not code length. Both recursive and iterative methods visit each node once, so time grows the same way.

Interview Connect

Understanding this recursive approach shows you can think about problems in different ways and handle linked lists confidently, a key skill in many coding challenges.

Self-Check

"What if we changed the recursion to an iterative loop? How would the time complexity change?"

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