Reverse a Singly Linked List Recursive in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Reverse on a 3-Node List

What is the printed state of the linked list after calling the recursive reverse function on the list 1 -> 2 -> 3 -> null?

DSA C#

public class Node {
    public int val;
    public Node next;
    public Node(int val) { this.val = val; this.next = null; }
}

public class LinkedList {
    public Node head;

    public Node ReverseRecursive(Node node) {
        if (node == null || node.next == null) return node;
        Node newHead = ReverseRecursive(node.next);
        node.next.next = node;
        node.next = null;
        return newHead;
    }

    public void PrintList(Node node) {
        while (node != null) {
            System.Console.Write(node.val + " -> ");
            node = node.next;
        }
        System.Console.WriteLine("null");
    }
}

// Setup
LinkedList list = new LinkedList();
list.head = new Node(1);
list.head.next = new Node(2);
list.head.next.next = new Node(3);

// Reverse
list.head = list.ReverseRecursive(list.head);
list.PrintList(list.head);

A 1 -> 2 -> 3 -> null

B null

C 3 -> 2 -> 1 -> null

D 3 -> 1 -> 2 -> null

💡 Hint

Think about how the recursion reverses the links starting from the end of the list.

The recursive function reverses the list by reaching the last node and then rewiring the next pointers backward. The final list is reversed: 3 -> 2 -> 1 -> null.

❓ Predict Output

intermediate

Output After Recursive Reverse on Single Node List

What is the printed state of the linked list after calling the recursive reverse function on a single-node list 5 -> null?

DSA C#

public class Node {
    public int val;
    public Node next;
    public Node(int val) { this.val = val; this.next = null; }
}

public class LinkedList {
    public Node head;

    public Node ReverseRecursive(Node node) {
        if (node == null || node.next == null) return node;
        Node newHead = ReverseRecursive(node.next);
        node.next.next = node;
        node.next = null;
        return newHead;
    }

    public void PrintList(Node node) {
        while (node != null) {
            System.Console.Write(node.val + " -> ");
            node = node.next;
        }
        System.Console.WriteLine("null");
    }
}

// Setup
LinkedList list = new LinkedList();
list.head = new Node(5);

// Reverse
list.head = list.ReverseRecursive(list.head);
list.PrintList(list.head);

A 5 -> null

B Error: NullReferenceException

C 5 -> 5 -> null

D null

💡 Hint

A single node reversed is itself.

The base case returns the node itself when next is null, so the list remains unchanged.

🔧 Debug

advanced

Identify the Error in Recursive Reverse Implementation

What error will this recursive reverse function cause when called on a list with multiple nodes?

DSA C#

public Node ReverseRecursive(Node node) {
    if (node == null) return null;
    Node newHead = ReverseRecursive(node.next);
    node.next.next = node;
    node.next = null;
    return newHead;
}

A NullReferenceException at node.next.next when node.next is null

B StackOverflowException due to infinite recursion

C No error, works correctly

D Compilation error due to missing return statement

💡 Hint

Check the base case carefully.

The base case is missing the condition to return when node.next is null, so node.next.next causes NullReferenceException.

🧠 Conceptual

advanced

Why Does Recursive Reverse Use node.next.next = node?

In the recursive reverse function, why do we assign 'node.next.next = node'?

A To set the current node's next pointer to null

B To skip the next node and link current node to the node after next

C To delete the next node from the list

D To make the next node point back to the current node, reversing the link

💡 Hint

Think about how links are reversed step by step.

Assigning 'node.next.next = node' reverses the direction of the link between current node and next node.

❓ Predict Output

expert

Output After Recursive Reverse on 4-Node List with Values 10, 20, 30, 40

What is the printed state of the linked list after calling the recursive reverse function on the list 10 -> 20 -> 30 -> 40 -> null?

DSA C#

public class Node {
    public int val;
    public Node next;
    public Node(int val) { this.val = val; this.next = null; }
}

public class LinkedList {
    public Node head;

    public Node ReverseRecursive(Node node) {
        if (node == null || node.next == null) return node;
        Node newHead = ReverseRecursive(node.next);
        node.next.next = node;
        node.next = null;
        return newHead;
    }

    public void PrintList(Node node) {
        while (node != null) {
            System.Console.Write(node.val + " -> ");
            node = node.next;
        }
        System.Console.WriteLine("null");
    }
}

// Setup
LinkedList list = new LinkedList();
list.head = new Node(10);
list.head.next = new Node(20);
list.head.next.next = new Node(30);
list.head.next.next.next = new Node(40);

// Reverse
list.head = list.ReverseRecursive(list.head);
list.PrintList(list.head);

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

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

C 40 -> 20 -> 30 -> 10 -> null

D Error: NullReferenceException

💡 Hint

The recursive reverse flips the list completely.

The recursive function reverses the list fully, so the last node becomes the head and links go backward.

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. Iterate through the list and reverse pointers in a loop

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

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's pointer after recursion

   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 D

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
• 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 new Node();

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

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

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

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 itself if it is null or the last node (head.next == null).

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Returning null when list has one node
• Accessing head.next without checking if head is null
• Creating new nodes unnecessarily

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. 3 -> 2 -> 1 -> null

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

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

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

Solution

1. Step 1: Trace recursion calls

   Reverse(1) calls Reverse(2), which calls Reverse(3). Reverse(3) returns node 3 as base case.

2. Step 2: Fix pointers on return

   head.next.next = head reverses link: 3.next = 2, then 2.next = 1, and 1.next = null.

3. Final Answer:

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

4. Quick Check:

   Recursive reversal output = reversed list [OK]

Hint: Follow recursion bottom-up to see pointer reversal [OK]

Common Mistakes:

• Not setting head.next to null causing cycle
• Confusing node values with pointers
• Returning wrong node as new head

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. Should return head instead of newHead

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

C. head.next.next = head causes null pointer; fix by checking head.next

D. No error; code is correct

Solution

1. Step 1: Check base case correctness

   Base case only checks head == null, missing single node case (head.next == null).

2. Step 2: Consequence of missing base case

   Without single node base case, head.next may be null causing head.next.next to throw error.

3. Final Answer:

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

4. Quick Check:

   Base case must handle empty and single node [OK]

Hint: Base case must cover empty and single node to avoid errors [OK]

Common Mistakes:

• Ignoring single node base case
• Not setting head.next = null after reversal
• Returning wrong node as new head

5. Given a singly linked list with nodes containing values 10 -> 20 -> 30 -> 40 -> null, what will be the printed output after applying this recursive reverse function?

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;
}

// After reversal, print list from newHead

hard

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

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

C. 40 -> 10 -> 20 -> 30 -> null

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

Solution

1. Step 1: Trace recursive reversal on given list

   Recursion reverses from last node 40 back to 10, fixing pointers step-by-step.

2. Step 2: Confirm final reversed list order

   After reversal, list order is 40 -> 30 -> 20 -> 10 -> null.

3. Final Answer:

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

4. Quick Check:

   Recursive reversal reverses entire list order [OK]

Hint: Reversal flips list order completely [OK]

Common Mistakes:

• Printing original list instead of reversed
• Forgetting to set head.next = null causing cycle
• Misreading pointer changes as value swaps