Reverse a Singly Linked List Recursive in DSA Rust - Build from Scratch

Reverse a Singly Linked List Recursively

📖 Scenario: You are working on a simple contact list app. The contacts are stored in a singly linked list, where each contact points to the next one. Sometimes, you want to see the contacts in reverse order.

🎯 Goal: Build a Rust program that creates a singly linked list of contacts, then reverses the list using a recursive function, and finally prints the reversed list.

📋 What You'll Learn

Create a singly linked list with exactly these contacts in order: "Alice", "Bob", "Charlie", "Diana"

Add a recursive function called reverse_list that reverses the linked list

Print the reversed linked list showing each contact's name followed by an arrow, ending with None

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like music playlists, undo functionality in apps, and managing tasks where order matters and dynamic insertion/removal is needed.

💼 Career

Understanding linked lists and recursion is fundamental for software engineers, especially for roles involving low-level programming, system design, and algorithm optimization.

1

Create the singly linked list

Create a singly linked list struct called Node with a field name of type String and a field next of type Option. Then create a variable called head that holds the list with these contacts in order: "Alice", "Bob", "Charlie", "Diana".

DSA Rust

// Define the Node struct and create the linked list with the given contacts
// Your code here

Hint

Define the Node struct with name and next. Then nest Box::new(Node { ... }) inside Some to build the list from "Alice" to "Diana".

2

Add a recursive reverse function

Add a recursive function called reverse_list that takes an Option called head and returns an Option. This function should reverse the linked list recursively and return the new head.

DSA Rust

struct Node {
    name: String,
    next: Option>,
}

fn reverse_list(head: Option>) -> Option> {
    // Your code here
}

fn main() {
    let head = Some(Box::new(Node {
        name: String::from("Alice"),
        next: Some(Box::new(Node {
            name: String::from("Bob"),
            next: Some(Box::new(Node {
                name: String::from("Charlie"),
                next: Some(Box::new(Node {
                    name: String::from("Diana"),
                    next: None,
                })),
            })),
        })),
    }));
}

Hint

Use pattern matching on head. If None or only one node, return it. Otherwise, recursively reverse the rest, then attach the current node at the end.

3

Add a function to print the linked list

Add a function called print_list that takes a reference to Option<Box<Node>> called head and prints the linked list names separated by -> and ending with None.

DSA Rust

struct Node {
    name: String,
    next: Option>,
}

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);
                if let Some(mut next_node) = new_head.as_deref_mut() {
                    let mut tail = &mut next_node;
                    while let Some(ref mut next_tail) = tail.next {
                        tail = next_tail;
                    }
                    tail.next = Some(node);
                    tail.next.as_mut().unwrap().next = None;
                }
                new_head
            }
        }
    }
}

fn print_list(head: &Option>) {
    // Your code here
}

fn main() {
    let head = Some(Box::new(Node {
        name: String::from("Alice"),
        next: Some(Box::new(Node {
            name: String::from("Bob"),
            next: Some(Box::new(Node {
                name: String::from("Charlie"),
                next: Some(Box::new(Node {
                    name: String::from("Diana"),
                    next: None,
                })),
            })),
        })),
    }));
}

Hint

Use a while let loop to traverse the list. Print each name followed by -> . After the loop, print None.

4

Reverse the list and print the result

In the main function, call reverse_list with head and store the result in a variable called reversed_head. Then call print_list with &reversed_head to print the reversed linked list.

DSA Rust

struct Node {
    name: String,
    next: Option>,
}

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);
                if let Some(mut next_node) = new_head.as_deref_mut() {
                    let mut tail = &mut next_node;
                    while let Some(ref mut next_tail) = tail.next {
                        tail = next_tail;
                    }
                    tail.next = Some(node);
                    tail.next.as_mut().unwrap().next = None;
                }
                new_head
            }
        }
    }
}

fn print_list(head: &Option>) {
    let mut current = head;
    while let Some(node) = current {
        print!("{} -> ", node.name);
        current = &node.next;
    }
    println!("None");
}

fn main() {
    let head = Some(Box::new(Node {
        name: String::from("Alice"),
        next: Some(Box::new(Node {
            name: String::from("Bob"),
            next: Some(Box::new(Node {
                name: String::from("Charlie"),
                next: Some(Box::new(Node {
                    name: String::from("Diana"),
                    next: None,
                })),
            })),
        })),
    }));
    // Your code here
}

Hint

Call reverse_list(head) and save it in reversed_head. Then call print_list(&reversed_head) to show the reversed list.

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