Reverse a Singly Linked List Recursive in DSA Go - 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 represents a singly linked list node
type Node struct {
    Next *Node
}

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

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.
• 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 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 number of operations 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 because it uses recursion instead of loops."

[OK] Correct: Both recursion and loops visit each node once, so they take similar time. Recursion adds function call overhead but does not reduce the number of steps.

Interview Connect

Understanding recursive linked list reversal helps you think clearly about recursion and linked structures, a skill useful in many coding challenges.

Self-Check

"What if we changed the recursive reversal to an iterative approach? How would the time complexity change?"

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