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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the base case for the recursive reverse function.

DSA Java

if (head == [1] || head.next == null) {
    return head;
}

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

Anull

Bthis

Chead.next

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using head instead of null for the base case.

Checking head.next.next instead of head.next.

The base case returns the head when it is null or the next node is null, meaning the list is empty or has one node.

2fill in blank

medium

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

DSA Java

ListNode newHead = [1](head.next);

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AreverseList

BreverseListRecursive

Creverse

DreverseLinkedList

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different function name than the one defined.

Calling reverseList with head instead of head.next.

The recursive call is made to reverseList to reverse the list starting from head.next.

3fill in blank

hard

Fix the error in linking the next node's next pointer to the current head.

DSA Java

head.next.[1] = head;

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Aprev

Bnext

CnextNode

Dpointer

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev or pointer which are not defined.

Using head.next.prev which does not exist.

We set head.next.next to head to reverse the link between nodes.

4fill in blank

hard

Fill both blanks to complete the code that breaks the old link and returns the new head.

DSA Java

head.next = [1];
return [2];

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Anull

BnewHead

Chead

Dhead.next

Attempts:

3 left

💡 Hint

Common Mistakes

Returning head instead of newHead.

Not setting head.next to null causing cycles.

We set head.next to null to break the old link and return newHead as the new head of the reversed list.

5fill in blank

hard

Fill all three blanks to complete the recursive reverse function.

DSA Java

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

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

Anull

BnewHead

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names or pointers.

Forgetting to set head.next to null.

The base case checks for null, newHead stores the reversed list, and head.next.next points back to head to reverse the link.

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