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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the base case node in the recursive reverse function.

DSA Python

def reverse_recursive(node):
    if node is None or node.next is None:
        return [1]

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

Anode

Bnode.next

CNone

Dnode.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Returning node.next causes None to be returned incorrectly.

Returning node.prev is invalid for singly linked list.

The base case returns the current node when it is the last node or None.

2fill in blank

medium

Complete the code to recursively reverse the rest of the list after the current node.

DSA Python

def reverse_recursive(node):
    if node is None or node.next is None:
        return node
    new_head = [1]

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

Areverse_recursive(node.prev)

Breverse_recursive(node)

Creverse_recursive(node.next)

Dreverse_recursive(None)

Attempts:

3 left

💡 Hint

Common Mistakes

Calling reverse_recursive(node) causes infinite recursion.

Calling reverse_recursive(node.prev) is invalid for singly linked list.

We call reverse_recursive on node.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 Python

def reverse_recursive(node):
    if node is None or node.next is None:
        return node
    new_head = reverse_recursive(node.next)
    node.next.next = [1]
    node.next = None
    return new_head

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

Anew_head

Bnode

CNone

Dnode.next

Attempts:

3 left

💡 Hint

Common Mistakes

Setting node.next.next to None breaks the chain.

Setting node.next.next to new_head causes incorrect links.

We set node.next.next to node to reverse the link direction.

4fill in blank

hard

Fill both blanks to complete the recursive reversal and break the old link.

DSA Python

def reverse_recursive(node):
    if node is None or node.next is None:
        return node
    new_head = reverse_recursive([1])
    [2].next.next = node
    node.next = None
    return new_head

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

Anode.next

Bnode

Cnew_head

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Using new_head in place of node.next causes errors.

Setting node.next.next to new_head breaks the link.

We recurse on node.next and set node.next.next to node to reverse the link.

5fill in blank

hard

Fill all three blanks to complete the recursive function that reverses a singly linked list.

DSA Python

def reverse_recursive(node):
    if node is None or node.next is None:
        return [1]
    new_head = reverse_recursive([2])
    [3].next.next = node
    node.next = None
    return new_head

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

Anode

Bnode.next

Dnew_head

Attempts:

3 left

💡 Hint

Common Mistakes

Returning node.next in base case is wrong.

Using new_head instead of node.next for link reversal causes errors.

The base case returns node. We recurse on node.next and set node.next.next to node to reverse the link.

Practice

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

easy

A. It iterates through the list using a loop to reverse pointers.

B. It swaps the values of the first and last nodes only.

C. It reverses the rest of the list and then adjusts the current node's next pointer.

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 end (base case).

2. Step 2: Reverse pointers on return

   After reaching the end, it reverses the next pointer of the current node to point backward.

3. Final Answer:

   It reverses the rest of the list and then adjusts the current node's next pointer. -> Option C

4. Quick Check:

   Recursive reversal = reverse rest + fix pointers [OK]

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

Common Mistakes:

• Trying to swap node values instead of pointers
• Using loops instead of recursion
• Not handling the base case properly

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

easy

A. if head.next is None: return None

B. if head.next is None: return head.next

C. if head is None: return head.next

D. if head is None or head.next is None: return head

Solution

1. Step 1: Identify base case meaning

   The base case stops recursion when the list is empty or has one node.

2. Step 2: Check correct condition

   When head is None (empty) or head.next is None (single node), return head as reversed list.

3. Final Answer:

   if head is None or head.next is None: return head -> Option D

4. Quick Check:

   Base case = empty or single node [OK]

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

Common Mistakes:

• Returning None incorrectly for single node
• Missing the empty list check
• Returning head.next instead of head

3. What will be the output linked list after reversing recursively the list: 1 -> 2 -> 3 -> None?

medium

A. 1 -> 2 -> 3 -> None

B. 3 -> 2 -> 1 -> None

C. 2 -> 3 -> 1 -> None

D. None -> 3 -> 2 -> 1

Solution

1. Step 1: Trace recursive calls

   Function calls itself on 2 -> 3 -> None, then on 3 -> None (base case).

2. Step 2: Reverse pointers on return

   On return, 3 points to 2, 2 points to 1, and 1 points to None.

3. Final Answer:

   3 -> 2 -> 1 -> None -> Option B

4. Quick Check:

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

Hint: Last node becomes new head after recursion [OK]

Common Mistakes:

• Not updating the last node's next pointer
• Returning original head instead of new head
• Confusing node values with pointers

4. Given this recursive reverse function, what is the error?

def reverse(head):
    if head is None:
        return None
    new_head = reverse(head.next)
    head.next.next = head
    head.next = None
    return new_head

medium

A. Base case misses single node check causing AttributeError

B. head.next.next = head causes infinite loop

C. head.next = None should be before recursive call

D. Function returns None always

Solution

1. Step 1: Analyze base case

   Base case only checks if head is None, but not if head.next is None (single node).

2. Step 2: Identify error cause

   When head.next is None, head.next.next causes AttributeError because head.next is None.

3. Final Answer:

   Base case misses single node check causing AttributeError -> Option A

4. Quick Check:

   Missing single node base case = error [OK]

Hint: Always check head.next is None in base case [OK]

Common Mistakes:

• Ignoring single node base case
• Changing pointers before recursion
• Returning None instead of new head

5. How would you modify the recursive reverse function to also return the tail node of the reversed list efficiently?

hard

A. Return a tuple (new_head, tail) and update tail during recursion

B. Use a global variable to track tail node

C. Traverse the reversed list again to find tail after recursion

D. Return only new_head and ignore tail

Solution

1. Step 1: Understand tail tracking need

   To get tail efficiently, return it along with new head during recursion.

2. Step 2: Modify function to return tuple

   Return (new_head, tail) from recursive calls and update tail as current node.

3. Step 3: Avoid extra traversal

   This avoids traversing reversed list again, improving efficiency.

4. Final Answer:

   Return a tuple (new_head, tail) and update tail during recursion -> Option A

5. Quick Check:

   Return (head, tail) tuple for efficiency [OK]

Hint: Return both head and tail as tuple in recursion [OK]

Common Mistakes:

• Using global variables causing side effects
• Traversing list again after reversal
• Ignoring tail node requirement