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 grows as the list gets longer.

Specifically, how many steps does the recursive method take as the list size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct Node {
    int data;
    struct Node* next;
};

struct Node* reverseRecursive(struct Node* head) {
    if (head == NULL || head->next == NULL) return head;
    struct Node* rest = reverseRecursive(head->next);
    head->next->next = head;
    head->next = NULL;
    return rest;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

Each node causes one recursive call, so the number of steps grows 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: "Recursive reversal is faster than iterative reversal because recursion is magical."

[OK] Correct: Both recursive and iterative methods visit each node once, so they take similar time. Recursion adds extra overhead from function calls but does not reduce the total steps.

Interview Connect

Understanding recursive reversal helps you think about how recursion works on linked structures, a key skill for many coding problems.

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. Use a stack to store nodes and pop them in reverse order

B. Reverse the rest of the list and then fix the first node's next pointer

C. Swap the data values of nodes from start to end

D. Iterate through the list and reverse pointers one by one

Solution

1. Step 1: Understand recursion on linked list

   Recursion reverses the list from the second node onward first.

2. Step 2: Fix the first node's next pointer

   After recursion returns, the first node's next pointer is set to null and the previous node points back to it.

3. Final Answer:

   Reverse the rest of the list and then fix the first node's next pointer -> Option B

4. Quick Check:

   Recursive reversal = Reverse rest + fix first [OK]

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

Common Mistakes:

• Trying to swap node data instead of pointers
• Not setting the first 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 == NULL) return NULL;

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

C. if (head == NULL || head->next == NULL) return head;

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

Solution

1. Step 1: Identify base cases for recursion

   Base case must handle empty list and single node list to stop recursion.

2. Step 2: Choose correct condition

   Condition 'head == NULL || head->next == NULL' covers both empty and single node cases.

3. Final Answer:

   if (head == NULL || head->next == NULL) return head; -> Option C

4. Quick Check:

   Base case = empty or single node [OK]

Hint: Base case stops recursion on empty or single node [OK]

Common Mistakes:

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

3. What will be the output linked list after reversing recursively the list: 1 -> 2 -> 3 -> NULL?

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 recursive reversal

   Recursion reverses from last node back: 3 becomes new head, 2 points to 1, 1 points to NULL.

2. Step 2: Confirm final list order

   Final reversed list is 3 -> 2 -> 1 -> NULL.

3. Final Answer:

   3 -> 2 -> 1 -> NULL -> Option D

4. Quick Check:

   Reverse(1->2->3) = 3->2->1 [OK]

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

Common Mistakes:

• Assuming original order remains
• Mixing node values instead of pointers
• Forgetting to set last node's next to NULL

4. Identify the error in this recursive reverse function snippet:

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

medium

A. head->next is set to head instead of NULL causing a cycle

B. Missing base case for empty list

C. Incorrect return value; should return head

D. No recursive call made

Solution

1. Step 1: Analyze pointer assignments

   head->next->next = head correctly points next node back to current node.

2. Step 2: Identify incorrect pointer reset

   head->next = head wrongly points current node's next to itself, causing a cycle instead of NULL.

3. Final Answer:

   head->next is set to head instead of NULL causing a cycle -> Option A

4. Quick Check:

   head->next must be NULL, not head [OK]

Hint: Set head->next to NULL to avoid cycles [OK]

Common Mistakes:

• Setting head->next to head causing infinite loop
• Forgetting to return the new head
• Not handling base case properly

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

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

void printList(Node* head) {
  while (head) {
    printf("%d -> ", head->data);
    head = head->next;
  }
  printf("NULL\n");
}

int main() {
  // List creation omitted for brevity
  Node* reversed = reverse(head);
  printList(reversed);
}

hard

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

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

C. NULL

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

Solution

1. Step 1: Trace recursive reversal on list 10->20->30->40

   Recursion reverses from last node: 40 becomes new head, pointers reversed stepwise.

2. Step 2: Confirm print output

   Print function outputs nodes in order: 40 -> 30 -> 20 -> 10 -> NULL.

3. Final Answer:

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

4. Quick Check:

   Reverse(10->20->30->40) = 40->30->20->10 [OK]

Hint: Last node becomes head; print from new head [OK]

Common Mistakes:

• Assuming original order remains after reverse
• Forgetting to set last node's next to NULL
• Confusing print output with input list