Reverse a Singly Linked List Recursive in DSA C++ - Time & Space Complexity

Time Complexity: Reverse a Singly Linked List Recursive

O(n)

Understanding Time Complexity

We want to understand how the time needed to reverse a linked list using recursion changes as the list grows.

How does the number of steps grow when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code reverses a singly linked list by calling itself on the next node until it reaches the end, then rewires the pointers backward.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive call on the next node.
• How many times: Once for each node in the list, so n times for n nodes.

How Execution Grows With Input

Each node causes one recursive call, so the total steps grow directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 recursive calls and rewiring steps
100	About 100 recursive calls and rewiring steps
1000	About 1000 recursive calls and rewiring steps

Pattern observation: The work grows linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows in direct proportion to the number of nodes.

Common Mistake

[X] Wrong: "Recursion makes this algorithm slower than iterative because it repeats work multiple times."

[OK] Correct: Each node is visited exactly once in the recursion, so no repeated work happens. The recursion just adds function call overhead but does not increase the number of steps beyond n.

Interview Connect

Understanding recursive reversal shows you can think about problems in different ways and handle linked lists confidently, a skill often valued in interviews.

Self-Check

"What if we changed the recursion to an iterative loop? How would the time complexity change?"

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