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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the base case for the recursive reverse function.

DSA Rust

fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    if head.is_none() || head.as_ref().unwrap().next.is_none() {
        return [1];
    }
    // recursive case omitted
    None
}

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

Ahead.clone()

BNone

Chead.unwrap()

DSome(head)

Attempts:

3 left

💡 Hint

Common Mistakes

Returning None instead of head for base case

Trying to unwrap head without checking

The base case returns the head itself when the list is empty or has one node, so we return a clone of head.

2fill in blank

medium

Complete the code to recursively reverse the rest of the list after the head.

DSA Rust

fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    if head.is_none() || head.as_ref().unwrap().next.is_none() {
        return head.clone();
    }
    let mut head = head.unwrap();
    let rest = head.next.take();
    let reversed = [1];
    // rest of code omitted
    None
}

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Areverse_list(rest)

Breverse_list(head.next)

Creverse_list(Some(head))

Dreverse_list(None)

Attempts:

3 left

💡 Hint

Common Mistakes

Calling reverse_list on head.next without taking ownership

Passing None instead of rest

We recursively call reverse_list on rest, which is the next part of the list after head.

3fill in blank

hard

Fix the error in linking the reversed list back to the current head node.

DSA Rust

fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    if head.is_none() || head.as_ref().unwrap().next.is_none() {
        return head.clone();
    }
    let mut head = head.unwrap();
    let rest = head.next.take();
    let reversed = reverse_list(rest);
    [1];
    head.next = None;
    reversed
}

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Ahead.next.as_mut().unwrap().next = Some(head);

Bhead.next = reversed;

Creversed.as_mut().unwrap().next = Some(head);

Dreversed.unwrap().next = Some(head);

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to access head.next when it is None

Assigning to head.next instead of reversed list

We must set the next of the last node in reversed list to head, so we use reversed.as_mut().unwrap().next = Some(head).

4fill in blank

hard

Fill both blanks to correctly complete the recursive reverse function.

DSA Rust

fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    if head.is_none() || head.as_ref().unwrap().next.is_none() {
        return [1];
    }
    let mut head = head.unwrap();
    let rest = head.next.take();
    let reversed = reverse_list(rest);
    [2];
    head.next = None;
    reversed
}

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

Ahead.clone()

Breversed.as_mut().unwrap().next = Some(head)

Chead.next = reversed

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Returning None in base case

Incorrectly linking nodes in recursive step

Return head.clone() in base case and link reversed list's last node to head with reversed.as_mut().unwrap().next = Some(head).

5fill in blank

hard

Fill all three blanks to implement the recursive reverse function correctly.

DSA Rust

fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
    if [1] || head.as_ref().unwrap().next.is_none() {
        return head.clone();
    }
    let mut head = head.unwrap();
    let rest = head.next.take();
    let reversed = reverse_list(rest);
    [2];
    head.next = [3];
    reversed
}

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

Ahead.is_none()

Breversed.as_mut().unwrap().next = Some(head)

CNone

Dhead.is_some()

Attempts:

3 left

💡 Hint

Common Mistakes

Using head.is_some() in base case condition

Forgetting to set head.next to None

Check if head is None for base case, link reversed list's last node to head, and set head.next to None to avoid cycles.

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 adjust the current node's next pointer

B. Swap the values 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 recursive reversal logic

   In recursion, we reverse the list starting from the end by calling the function on the next node first.

2. Step 2: Adjust pointers after recursion returns

   After the recursive call returns, we set the next node's next pointer to the current node, effectively reversing the link.

3. Final Answer:

   Reverse the rest of the list first, then adjust the current node's next pointer -> Option A

4. Quick Check:

   Recursive reversal = reverse rest first [OK]

Hint: Think: reverse tail first, then fix current link [OK]

Common Mistakes:

• Trying to swap node values instead of pointers
• Reversing pointers before recursive call
• Using iteration instead of recursion

2. Which of the following Rust function signatures correctly represents a recursive function to reverse a singly linked list?

easy

A. fn reverse_list(head: Node) -> Node

B. fn reverse_list(head: &mut Node) -> &mut Node

C. fn reverse_list(head: Option) -> Option

D. fn reverse_list(head: &Node) -> Node

Solution

1. Step 1: Identify correct ownership and return type

   In Rust, linked lists use Option to represent nullable pointers with ownership.

2. Step 2: Match function signature for recursive reversal

   The function should take ownership of the list head and return the new head after reversal, so Option input and output is correct.

3. Final Answer:

   fn reverse_list(head: Option) -> Option -> Option C

4. Quick Check:

   Ownership with Option = fn reverse_list(head: Option) -> Option [OK]

Hint: Use Option for linked list nodes in Rust [OK]

Common Mistakes:

• Using references without ownership causing borrow errors
• Returning plain Node instead of Option
• Passing mutable references incorrectly

3. Given the following Rust code snippet for recursive reversal, what will be the output list after reversal if the input list is 1 -> 2 -> 3 -> null?

fn reverse_list(head: Option) -> Option {
    match head {
        None => None,
        Some(mut node) => {
            if node.next.is_none() {
                Some(node)
            } else {
                let next = node.next.take();
                let new_head = reverse_list(next);
                let mut tail = new_head.as_ref();
                while let Some(t) = tail {
                    if t.next.is_none() {
                        break;
                    }
                    tail = t.next.as_ref();
                }
                tail.unwrap().next = Some(node);
                node.next = None;
                new_head
            }
        }
    }
}

medium

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. 2 -> 3 -> 1 -> null

D. Compile-time error due to borrowing

Solution

1. Step 1: Analyze pointer manipulation and borrowing

   The code tries to get a mutable reference to tail via as_ref() and then mutate tail.unwrap().next, which causes a borrowing error in Rust.

2. Step 2: Understand Rust's borrowing rules

   Rust does not allow mutable access through an immutable reference (as_ref()), so this code will cause a compile-time error.

3. Final Answer:

   Compile-time error due to borrowing -> Option D

4. Quick Check:

   Immutable ref then mutable write = error [OK]

Hint: Check mutable vs immutable references in Rust code [OK]

Common Mistakes:

• Ignoring Rust's strict borrowing rules
• Assuming as_ref() allows mutation
• Not using take() properly to avoid borrow conflicts

4. Identify the bug in this recursive Rust function to reverse a singly linked list:

fn reverse_list(head: Option) -> Option {
    match head {
        None => None,
        Some(mut node) => {
            if node.next.is_none() {
                Some(node)
            } else {
                let next = node.next.take();
                let new_head = reverse_list(next);
                node.next.as_mut().unwrap().next = Some(node);
                node.next = None;
                new_head
            }
        }
    }
}

medium

A. Function does not return the new head

B. Using node.next.as_mut().unwrap() after take() causes None unwrap panic

C. Missing base case for empty list

D. No bug, code is correct

Solution

1. Step 1: Understand effect of take() on node.next

   Calling node.next.take() sets node.next to None, so node.next.as_mut() returns None.

2. Step 2: Identify unsafe unwrap causing panic

   Calling unwrap() on None causes a runtime panic, so node.next.as_mut().unwrap() is invalid after take().

3. Final Answer:

   Using node.next.as_mut().unwrap() after take() causes None unwrap panic -> Option B

4. Quick Check:

   unwrap on None after take() = panic [OK]

Hint: Don't unwrap after take() without checking for None [OK]

Common Mistakes:

• Assuming node.next still exists after take()
• Ignoring unwrap panic possibility
• Not returning new_head properly

5. You want to reverse a singly linked list recursively in Rust, but also count how many nodes were reversed. Which approach correctly combines reversal and counting? Options: A) Modify the recursive function to return a tuple (new_head, count) and increment count after each recursion. B) Use a global mutable variable to count nodes during recursion. C) Reverse the list first, then iterate again to count nodes. D) Count nodes before reversal and ignore during recursion.

hard

A. Modify the recursive function to return a tuple (new_head, count) and increment count after each recursion.

B. Use a global mutable variable to count nodes during recursion.

C. Reverse the list first, then iterate again to count nodes.

D. Count nodes before reversal and ignore during recursion.

Solution

1. Step 1: Combine reversal and counting in recursion

   Modify the recursive function to return both the new head and the count of nodes reversed as a tuple.

2. Step 2: Increment count after each recursive call

   Each recursive call returns the count from deeper calls; add 1 for the current node before returning.

3. Final Answer:

   Modify the recursive function to return a tuple (new_head, count) and increment count after each recursion. -> Option A

4. Quick Check:

   Return tuple (head, count) in recursion = Modify the recursive function to return a tuple (new_head, count) and increment count after each recursion. [OK]

Hint: Return (head, count) tuple from recursion to track nodes [OK]

Common Mistakes:

• Using global variables causing unsafe code
• Counting nodes separately after reversal wastes time
• Ignoring counting during recursion