Reverse a Singly Linked List Recursive in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

if (head == [1] || head->next == nullptr) {
    return head;
}

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Anullptr

Bhead

CNULL

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of nullptr

Using head instead of nullptr for empty list

The base case for recursion is when the list is empty (head == nullptr) or has only one node (head->next == nullptr). We return head in these cases.

2fill in blank

medium

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

DSA C++

ListNode* newHead = [1](head->next);

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AreverseLinkedList

Breverse

CreverseList

DreverseRec

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong function name

Calling on head instead of head->next

The function calls itself recursively on the next node to reverse the rest of the list.

3fill in blank

hard

Fix the error in the pointer reversal step to correctly reverse the link.

DSA C++

head->next->[1] = head;

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Aprev

Bnext

Cnext->next

Dnext->prev

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev which does not exist in singly linked list

Using next->next which is incorrect

We set the next node's next pointer to point back to the current head to reverse the link.

4fill in blank

hard

Fill both blanks to complete the recursive reversal by breaking the old link and returning the new head.

DSA C++

head->next = [1];
return [2];

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Anullptr

BnewHead

Chead

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Returning head instead of newHead

Setting head->next to head causing cycle

We set head->next to nullptr to break the old link and return newHead which is the head of the reversed list.

5fill in blank

hard

Fill all three blanks to complete the full recursive reverse function.

DSA C++

ListNode* reverseList(ListNode* head) {
    if (head == [1] || head->next == nullptr) {
        return head;
    }
    ListNode* newHead = reverseList(head->next);
    head->next->[2] = head;
    head->next = [3];
    return newHead;
}

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

Anullptr

Bnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev which does not exist

Not breaking the old link causing cycles

The base case checks for nullptr. Then we reverse the link by setting head->next->next = head, break old link by head->next = nullptr, and return newHead.

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 data 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 elements 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's pointer after recursion

   After reversing the rest, the current node's next node should point back to current node, then current node points to null.

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 first [OK]

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

Common Mistakes:

• Trying to swap data instead of pointers
• Not setting current node's next to null
• Confusing iterative and recursive methods

2. Which of the following is the correct base case for a recursive function that reverses a singly linked list?

easy

A. if (head->next == nullptr) return head;

B. if (head == nullptr || head->next == nullptr) return head;

C. if (head == nullptr) return nullptr;

D. if (head->next == nullptr) head = nullptr;

Solution

1. Step 1: Identify base cases for recursion

   Base case should handle empty list and single node list, both are already reversed.

2. Step 2: Choose condition covering both cases

   Condition 'head == nullptr || head->next == nullptr' returns head for empty or single node list.

3. Final Answer:

   if (head == nullptr || head->next == nullptr) return head; -> Option B

4. Quick Check:

   Base case = empty or single node [OK]

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

Common Mistakes:

• Returning nullptr for single node list
• Not handling empty list
• Incorrectly modifying head in base case

3. Given the following linked list: 1 -> 2 -> 3 -> null, what will be the printed output after applying this recursive reverse function?

Node* reverse(Node* head) {
  if (!head || !head->next) return head;
  Node* newHead = reverse(head->next);
  head->next->next = head;
  head->next = nullptr;
  return newHead;
}

medium

A. 3 -> 1 -> 2 -> null

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

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

D. 3 -> 2 -> 1 -> 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: Reverse pointers on return

   On unwind from reverse(3), in reverse(2): head(2)->next(3)->next = 2 so 3->next=2, then 2->next=nullptr. Then in reverse(1): 2->next=1, 1->next=nullptr.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option D

4. Quick Check:

   Reversed list = 3->2->1 [OK]

Hint: Follow recursion unwind to see pointer reversal [OK]

Common Mistakes:

• Not setting head->next to nullptr causing cycle
• Confusing node values order
• Returning wrong node as new head

4. Identify the bug in this recursive reverse function and select the correct fix:

Node* reverse(Node* head) {
  if (!head || !head->next) return head;
  Node* newHead = reverse(head->next);
  head->next->next = head;
  return newHead;
}

medium

A. Remove the recursive call and use iteration

B. Change base case to 'if (head->next == nullptr) return head;'

C. Add 'head->next = nullptr;' before returning newHead

D. Return 'head' instead of 'newHead'

Solution

1. Step 1: Analyze missing pointer update

   Function misses setting 'head->next = nullptr', causing cycle in reversed list.

2. Step 2: Fix by setting 'head->next = nullptr'

   Adding 'head->next = nullptr;' breaks old link, preventing cycle and correctly terminates list.

3. Final Answer:

   Add 'head->next = nullptr;' before returning newHead -> Option C

4. Quick Check:

   Break old next link to avoid cycle [OK]

Hint: Always set current node's next to nullptr after reversing [OK]

Common Mistakes:

• Forgetting to break old next pointer
• Incorrect base case causing null pointer dereference
• Returning wrong node as new head

5. You want to reverse a singly linked list recursively but also count how many nodes were reversed. Which modification correctly adds counting without breaking the reversal?

Node* reverseCount(Node* head, int &count) {
  if (!head || !head->next) {
    count++;
    return head;
  }
  Node* newHead = reverseCount(head->next, count);
  head->next->next = head;
  head->next = nullptr;
  count++;
  return newHead;
}

What will be the value of count after reversing a list with 4 nodes?

hard

A. 4

B. 3

C. 5

D. 0

Solution

1. Step 1: Understand counting in base case

   When recursion reaches last node, count increments by 1 for that node.

2. Step 2: Count increments on each return step

   Each return from recursion increments count by 1 for each node processed, total nodes counted equals list length.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Count increments once per node [OK]

Hint: Increment count once per node during recursion [OK]

Common Mistakes:

• Counting only in base case missing other nodes
• Incrementing count twice per node
• Not passing count by reference