Reverse a Singly Linked List Recursive in DSA Java - 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 using recursion changes as the list grows.

How does the number of steps grow when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public ListNode reverseList(ListNode head) {
    if (head == null || head.next == null) {
        return head;
    }
    ListNode 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 reverses the links on the way back.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive call on the next node of the list.
• How many times: Once for each node in the list, so 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 pointer changes
100	About 100 recursive calls and pointer changes
1000	About 1000 recursive calls and pointer changes

Pattern observation: The number of operations grows linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Recursive reversal is slower because it does extra work on each call like loops inside loops."

[OK] Correct: Each recursive call only processes one node and does a constant amount of work, so it does not multiply work like nested loops.

Interview Connect

Understanding recursive reversal shows you can work with linked lists and recursion clearly, a skill often checked in interviews.

Self-Check

"What if we changed the recursive reversal to an iterative approach? How would the time complexity change?"

Practice

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

easy

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

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

C. Iterate through the list and reverse pointers in a loop

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

Solution

1. Step 1: Understand recursion on linked list

   Recursion works by solving a smaller part first, here reversing from the next node onward.

2. Step 2: Fix current node after recursive call

   After reversing the rest, 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 A

4. Quick Check:

   Recursive reversal = Reverse rest then fix current [OK]

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

Common Mistakes:

• Trying to swap node values instead of pointers
• Reversing pointers before recursive call
• 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 null.next;

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

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

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

Solution

1. Step 1: Identify base case meaning

   The base case stops recursion when list is empty or has one node, so head is null or head.next is null.

2. Step 2: Check each option

   if (head == null || head.next == null) return head; correctly returns head when list is empty or single node. Others cause errors or wrong returns.

3. Final Answer:

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

4. Quick Check:

   Base case = empty or single node returns head [OK]

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

Common Mistakes:

• Returning null when list has one node
• Accessing next of null causing errors
• Missing base case leading to infinite recursion

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

Node reverse(Node head) {
  if (head == null || head.next == null) return head;
  Node newHead = reverse(head.next);
  head.next.next = head;
  head.next = null;
  return newHead;
}

medium

A. null

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

C. 3 -> 2 -> 1 -> null

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

Solution

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

   Recursion goes to last node 3, then reverses links backward: 3 points to 2, 2 points to 1.

2. Step 2: Final reversed list

   After fixing pointers, the list becomes 3 -> 2 -> 1 -> null.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option C

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 cycle
• Returning old head instead of new head
• Confusing node values with pointers

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

Node reverse(Node head) {
  if (head == null) return null;
  Node newHead = reverse(head.next);
  head.next.next = head;
  return newHead;
}

medium

A. head.next.next = head; should be head.next = head.next.next;

B. Missing base case for single node; add 'head.next == null' check

C. Return statement should be 'return head;' instead of newHead

D. No error; code is correct

Solution

1. Step 1: Check base case correctness

   Base case only checks head == null, missing single node case causes null pointer error at head.next.next.

2. Step 2: Fix base case

   Adding 'head.next == null' to base case prevents error and correctly stops recursion.

3. Final Answer:

   Missing base case for single node; add 'head.next == null' check -> Option B

4. Quick Check:

   Base case must handle single node [OK]

Hint: Base case must check single node to avoid null pointer [OK]

Common Mistakes:

• Ignoring single node base case causing runtime error
• Incorrect pointer assignment causing infinite loop
• Returning wrong node after recursion

5. Given a singly linked list with nodes 10 -> 20 -> 30 -> 40 -> null, you want to reverse it recursively but also count how many nodes were reversed. Which modification correctly adds counting nodes during recursion?

class Result {
  Node head;
  int count;
}

Result reverseCount(Node head) {
  if (head == null || head.next == null) {
    Result res = new Result();
    res.head = head;
    res.count = (head == null) ? 0 : 1;
    return res;
  }
  Result res = reverseCount(head.next);
  head.next.next = head;
  head.next = null;
  res.count += 1;
  return res;
}

hard

A. This code correctly reverses and counts nodes

B. Count should be incremented before recursive call

C. head.next = null; should be removed to avoid breaking list

D. Result class should store count as static variable

Solution

1. Step 1: Analyze recursive counting logic

   Base case sets count to 1 or 0. Each recursive return increments count by 1 after reversing pointer.

2. Step 2: Confirm correctness of pointer reversal and counting

   Pointer reversal is correct and count increments after recursion, so total nodes counted correctly.

3. Final Answer:

   This code correctly reverses and counts nodes -> Option A

4. Quick Check:

   Counting nodes during recursion works as coded [OK]

Hint: Increment count after recursion to count all nodes [OK]

Common Mistakes:

• Incrementing count before recursion causing wrong totals
• Removing head.next = null causing cycles
• Using static variable causing shared state bugs