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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Linked List Reversal

What is the printed state of the linked list after reversing it recursively?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def print_list(head):
    current = head
    result = []
    while current:
        result.append(str(current.val))
        current = current.next
    print(' -> '.join(result) + ' -> null')

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

# Create linked list 1 -> 2 -> 3 -> 4 -> null
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)

head = reverse_recursive(head)
print_list(head)

A null

B 1 -> 2 -> 3 -> 4 -> null

C 4 -> 3 -> 2 -> 1 -> null

D 1 -> null

💡 Hint

Think about how the recursion unwinds and reverses the links one by one.

The recursive function reverses the linked list by moving to the end and then changing the next pointers backward. The final printed list is reversed.

❓ Predict Output

intermediate

Output when reversing a single-node list recursively

What is the printed state of the linked list after reversing a single-node list recursively?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def print_list(head):
    current = head
    result = []
    while current:
        result.append(str(current.val))
        current = current.next
    print(' -> '.join(result) + ' -> null')

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

# Create linked list 5 -> null
head = Node(5)

head = reverse_recursive(head)
print_list(head)

A 5 -> null

B null

C 5 -> 5 -> null

D Error: NoneType has no attribute 'next'

💡 Hint

A single-node list reversed is the same list.

The base case returns the head itself when the list has one node, so the output remains the same.

❓ Predict Output

advanced

Output of recursive reversal on empty list

What is the printed output when reversing an empty linked list recursively?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def print_list(head):
    if head is None:
        print('null')
        return
    current = head
    result = []
    while current:
        result.append(str(current.val))
        current = current.next
    print(' -> '.join(result) + ' -> null')

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

head = None
head = reverse_recursive(head)
print_list(head)

A Error: 'NoneType' object has no attribute 'next'

B null

C Empty list

D None

💡 Hint

Check the base case for empty list in the recursive function.

The function returns None for empty list and print_list prints 'null' for None head.

🔧 Debug

advanced

Identify the error in this recursive reversal code

What error will this code produce when reversing a linked list recursively?

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

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

# Create linked list 1 -> 2 -> null
head = Node(1)
head.next = Node(2)

head = reverse_recursive(head)

A RecursionError: maximum recursion depth exceeded

B AttributeError: 'NoneType' object has no attribute 'next'

C No error, works correctly

D TypeError: unsupported operand type(s)

💡 Hint

Check the line where the next pointer is assigned to itself.

The line 'head.next = head' creates a cycle causing infinite recursion leading to RecursionError.

🧠 Conceptual

expert

Why does the recursive reversal function set head.next.next = head?

In the recursive reversal of a singly linked list, why do we set 'head.next.next = head' before setting 'head.next = None'?

A To delete the current node from the list

B To break the link between current node and next node to avoid cycles

C To move the head pointer forward in the list

D To reverse the link by making the next node point back to the current node

💡 Hint

Think about how links are reversed step by step in recursion.

Setting 'head.next.next = head' reverses the direction of the link between two nodes, making the next node point back to the current node.

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