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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

func reverseList(head *ListNode) *ListNode {
    if head == nil || head.[1] == nil {
        return head
    }
    // recursive call
    return nil
}

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

Alink

Bprev

Cchild

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes a compile error because 'prev' is not a field.

Not checking for nil causes runtime errors.

The base case checks if the current node is nil or if its next pointer is nil, meaning it's the last node.

2fill in blank

medium

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

DSA Go

func reverseList(head *ListNode) *ListNode {
    if head == nil || head.next == nil {
        return head
    }
    newHead := reverseList(head.[1])
    return newHead
}

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

Achild

Bnext

Cprev

Dlink

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes compile errors.

Calling reverseList on head causes infinite recursion.

The recursive call is made on head.next to reverse the rest of the list.

3fill in blank

hard

Fix the error in the code to correctly reverse the link between nodes.

DSA Go

func reverseList(head *ListNode) *ListNode {
    if head == nil || head.next == nil {
        return head
    }
    newHead := reverseList(head.next)
    head.next.next = head
    head.[1] = nil
    return newHead
}

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

Anext

Bprev

Cchild

Dlink

Attempts:

3 left

💡 Hint

Common Mistakes

Setting head.prev causes compile errors.

Not setting head.next to nil causes infinite loops.

After reversing, head.next should be set to nil to avoid cycles.

4fill in blank

hard

Fill both blanks to complete the recursive reversal and return the new head.

DSA Go

func reverseList(head *ListNode) *ListNode {
    if head == nil || head.next == nil {
        return head
    }
    newHead := reverseList(head.[1])
    head.next.[2] = head
    head.next = nil
    return newHead
}

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

Anext

Bprev

Cchild

Dlink

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes errors.

Not setting head.next to nil causes cycles.

The recursive call is on head.next, and head.next.next is set to head to reverse the link.

5fill in blank

hard

Fill all three blanks to complete the recursive reverse function with proper pointer updates.

DSA Go

func reverseList(head *ListNode) *ListNode {
    if head == nil || head.[1] == nil {
        return head
    }
    newHead := reverseList(head.[2])
    head.next.[3] = head
    head.next = nil
    return newHead
}

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

Anext

Bprev

Cchild

Dlink

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' causes compile errors.

Not setting head.next to nil causes infinite loops.

All blanks use 'next' to correctly traverse and reverse the singly linked list.

Practice

1. What does the recursive function to reverse a singly linked list do at each step?

easy

A. It reverses the rest of the list and then attaches the current node at the end.

B. It swaps the first two nodes and stops.

C. It deletes the current node and moves to the next.

D. It creates a new list and copies nodes in reverse order.

Solution

1. Step 1: Understand recursion on linked list

   The function calls itself on the next node until it reaches the last node.

2. Step 2: Attach current node after reversing rest

   After recursion returns, it sets the next node's next pointer to current node, reversing the link.

3. Final Answer:

   It reverses the rest of the list and then attaches the current node at the end. -> Option A

4. Quick Check:

   Recursive reversal = attach current after reversed rest [OK]

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

Common Mistakes:

• Stopping recursion too early
• Not changing next pointers correctly
• Trying to reverse by swapping values

2. Which of the following is the correct base case for a recursive function reversing a singly linked list in Go?

easy

A. if head == nil { return nil }

B. if head == nil || head.Next == nil { return head }

C. if head.Next == nil { return nil }

D. if head == nil && head.Next == nil { return nil }

Solution

1. Step 1: Identify base case for recursion

   The recursion stops when the list is empty or has one node, so head is nil or head.Next is nil.

2. Step 2: Return head in base case

   Return the head node itself because a single node or empty list is already reversed.

3. Final Answer:

   if head == nil || head.Next == nil { return head } -> Option B

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 nil for single node
• Using && instead of || in condition
• Missing base case causing infinite recursion

3. What is the output linked list after running this recursive reverse function on the list 1 -> 2 -> 3 -> nil?

func reverse(head *Node) *Node {
    if head == nil || head.Next == nil {
        return head
    }
    newHead := reverse(head.Next)
    head.Next.Next = head
    head.Next = nil
    return newHead
}

medium

A. nil

B. 1 -> 2 -> 3 -> nil

C. 2 -> 3 -> 1 -> nil

D. 3 -> 2 -> 1 -> nil

Solution

1. Step 1: Trace recursion calls

   Calls reverse on 2 -> 3 -> nil, then on 3 -> nil (base case returns 3).

2. Step 2: Reverse links on return

   Sets 3.Next to 2, 2.Next to 1, and 1.Next to nil, reversing the list.

3. Final Answer:

   3 -> 2 -> 1 -> nil -> Option D

4. Quick Check:

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

Hint: Follow recursion bottom-up to see reversed links [OK]

Common Mistakes:

• Not setting head.Next to nil causing cycle
• Returning wrong node as new head
• Confusing order of pointer assignments

4. Identify the bug in this recursive reverse function for a singly linked list:

func reverse(head *Node) *Node {
    if head == nil {
        return nil
    }
    newHead := reverse(head.Next)
    head.Next.Next = head
    return newHead
}

medium

A. head.Next.Next should be head.Next

B. Incorrect return type

C. Missing base case for single node causing nil pointer dereference

D. No bug, code is correct

Solution

1. Step 1: Check base case correctness

   Base case only checks if head is nil, but not if head.Next is nil, causing head.Next.Next to panic if head.Next is nil.

2. Step 2: Identify nil pointer dereference

   When head.Next is nil, head.Next.Next causes runtime error because head.Next is nil.

3. Final Answer:

   Missing base case for single node causing nil pointer dereference -> Option C

4. Quick Check:

   Base case must handle single node to avoid nil panic [OK]

Hint: Always check base case for nil and single node [OK]

Common Mistakes:

• Ignoring single node base case
• Assuming head.Next is never nil
• Not setting head.Next to nil after reversal

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

func reverse(head *Node) *Node {
    if head == nil || head.Next == nil {
        return head
    }
    newHead := reverse(head.Next)
    head.Next.Next = head
    head.Next = nil
    return newHead
}

Choose the correct final linked list.

hard

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

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

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

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

Solution

1. Step 1: Recursively reverse from tail

   Recursion reaches node with value 40 (base case), then reverses links back up.

2. Step 2: Re-link nodes in reverse order

   Each node's next node points back to itself, and original head's next is set to nil.

3. Final Answer:

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

4. Quick Check:

   Reversed list = 40->30->20->10 [OK]

Hint: Reverse order means last node becomes new head [OK]

Common Mistakes:

• Forgetting to set head.Next to nil
• Returning old head instead of new head
• Misordering nodes after reversal