BST Find Maximum Element in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the maximum value in a BST by moving to the right child.

DSA Rust

fn find_max(node: &Option<Box<TreeNode>>) -> i32 {
    let mut current = node;
    while let Some(ref n) = current {
        if n.right.is_none() {
            return n.val;
        }
        current = &n.[1];
    }
    panic!("Empty tree");
}

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

Aright

Bleft

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using left child instead of right child.

Returning the root value without traversal.

To find the maximum element in a BST, we keep moving to the right child until there is no right child left.

2fill in blank

medium

Complete the code to handle the case when the BST is empty by returning None.

DSA Rust

fn find_max_option(node: &Option<Box<TreeNode>>) -> Option<i32> {
    let mut current = node;
    while let Some(ref n) = current {
        if n.right.is_none() {
            return Some(n.val);
        }
        current = &n.[1];
    }
    [2]
}

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

Aleft

Bright

CNone

DSome(0)

Attempts:

3 left

💡 Hint

Common Mistakes

Returning Some(0) for empty tree instead of None.

Using left child for traversal.

We move to the right child until none exists, then return the value wrapped in Some. If the tree is empty, return None.

3fill in blank

hard

Fix the error in the code to correctly find the maximum value recursively.

DSA Rust

fn find_max_recursive(node: &Option<Box<TreeNode>>) -> i32 {
    match node {
        None => panic!("Empty tree"),
        Some(ref n) => {
            if n.[1].is_none() {
                n.val
            } else {
                find_max_recursive(&n.right)
            }
        }
    }
}

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

Aparent

Bleft

Cval

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Checking left child instead of right child.

Returning node value without recursion.

The maximum value is found by recursively going to the right child until there is no right child.

4fill in blank

hard

Fill both blanks to create a function that returns the maximum value or None if the tree is empty.

DSA Rust

fn find_max_option_recursive(node: &Option<Box<TreeNode>>) -> Option<i32> {
    match node {
        None => [1],
        Some(ref n) => {
            if n.right.is_none() {
                Some(n.[2])
            } else {
                find_max_option_recursive(&n.right)
            }
        }
    }
}

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

ANone

BSome(0)

Cval

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Returning Some(0) for empty tree.

Accessing right instead of val for node value.

Return None if the tree is empty. Otherwise, return Some of the node's value when no right child exists.

5fill in blank

hard

Fill all three blanks to implement an iterative function that returns the maximum value or None if empty.

DSA Rust

fn find_max_iterative(node: &Option<Box<TreeNode>>) -> Option<i32> {
    let mut current = node;
    while let Some(ref n) = current {
        if n.[1].is_none() {
            return Some(n.[2]);
        }
        current = &n.[3];
    }
    None
}

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

Aleft

Bval

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using left child instead of right child.

Returning right child instead of val.

Check if right child is None to find max, return node's val, and move to right child in loop.

Practice

1. In a Binary Search Tree (BST), where is the maximum element always located?

easy

A. At any leaf node

B. At the leftmost node

C. At the rightmost node

D. At the root node

Solution

1. Step 1: Understand BST property

   In a BST, left child nodes are smaller and right child nodes are larger than the parent node.

2. Step 2: Locate maximum element

   The maximum element is the rightmost node because it has no larger right child.

3. Final Answer:

   At the rightmost node -> Option C

4. Quick Check:

   Maximum = rightmost node [OK]

Hint: Maximum is always the rightmost node in BST [OK]

Common Mistakes:

• Confusing maximum with root node
• Thinking maximum is at leftmost node
• Assuming maximum can be any leaf

2. Which Rust code snippet correctly finds the maximum element in a BST node structure?

easy

A. fn find_max(node: &Option) -> i32 { let mut current = node; while let Some(ref n) = current { if n.right.is_none() { return n.value; } current = &n.right; } unreachable!() }

B. fn find_max(node: &Option) -> i32 { node.unwrap().value }

C. fn find_max(node: &Option) -> i32 { node.as_ref().unwrap().left.unwrap().value }

D. fn find_max(node: &Option) -> i32 { 0 }

Solution

1. Step 1: Analyze code logic

   fn find_max(node: &Option) -> i32 { let mut current = node; while let Some(ref n) = current { if n.right.is_none() { return n.value; } current = &n.right; } unreachable!() } uses a loop to traverse right children until none remain, returning the rightmost node's value.

2. Step 2: Check other options

   fn find_max(node: &Option) -> i32 { node.unwrap().value } returns root value only, C returns left child value, D returns constant 0, all incorrect.

3. Final Answer:

   Iterative right traversal to rightmost node -> Option A

4. Quick Check:

   Traverse right until None = max value [OK]

Hint: Traverse right child until None to find max [OK]

Common Mistakes:

• Returning root value without traversal
• Accessing left child instead of right
• Returning a fixed value ignoring tree

3. What is the output of this Rust code that finds the maximum element in a BST?

struct Node { value: i32, left: Option, right: Option }

fn find_max(node: &Option) -> i32 {
    let mut current = node;
    while let Some(ref n) = current {
        if n.right.is_none() {
            return n.value;
        }
        current = &n.right;
    }
    unreachable!()
}

fn main() {
    let tree = Some(Box::new(Node {
        value: 10,
        left: Some(Box::new(Node { value: 5, left: None, right: None })),
        right: Some(Box::new(Node {
            value: 15,
            left: None,
            right: Some(Box::new(Node { value: 20, left: None, right: None }))
        }))
    }));
    println!("{}", find_max(&tree));
}

medium

A. 10

B. 20

C. 15

D. 5

Solution

1. Step 1: Trace the tree structure

   Root is 10, right child 15, right child's right child 20 (rightmost node).

2. Step 2: Follow find_max logic

   Function moves right until no right child: stops at node with value 20.

3. Final Answer:

   20 -> Option B

4. Quick Check:

   Rightmost node value = 20 [OK]

Hint: Rightmost node value is maximum [OK]

Common Mistakes:

• Returning root value instead of rightmost
• Confusing left and right children
• Not following loop to last right node

4. Identify the bug in this Rust function that tries to find the maximum element in a BST:

fn find_max(node: &Option) -> i32 {
    if node.is_none() {
        return 0;
    }
    let mut current = node;
    while current.is_some() {
        if current.as_ref().unwrap().right.is_none() {
            return current.as_ref().unwrap().value;
        }
        current = ¤t.as_ref().unwrap().left;
    }
    0
}

medium

A. It unwraps without checking None

B. It returns 0 for empty tree instead of error

C. It uses a loop instead of recursion

D. It traverses left child instead of right child

Solution

1. Step 1: Analyze traversal direction

   Function moves to left child in loop, but max is in right subtree.

2. Step 2: Identify correct traversal

   To find max, must traverse right child until None, not left.

3. Final Answer:

   It traverses left child instead of right child -> Option D

4. Quick Check:

   Max = rightmost node, not left [OK]

Hint: Max element is rightmost, so traverse right [OK]

Common Mistakes:

• Going left instead of right
• Returning 0 silently for empty tree
• Unwrapping without safe checks

5. Given a BST where some nodes have duplicate maximum values on the right subtree, how would you modify the Rust function to return the maximum value correctly?

hard

A. Traverse right until None and return the last node's value

B. Use recursion to check the left subtree and return max value

C. Traverse right and left children to find max value manually

D. Return the root value as maximum if duplicates exist

Solution

1. Step 1: Understand duplicates in BST

   In a BST (allowing duplicates typically on right), the maximum value(s) propagate to the rightmost node.

2. Step 2: Correct approach

   Traverse right until None; this reaches the rightmost node holding the maximum value, even with duplicates.

3. Final Answer:

   Traverse right until None and return the last node's value -> Option A

4. Quick Check:

   Rightmost = max even with duplicates [OK]

Hint: Right traversal finds max even with duplicates [OK]

Common Mistakes:

• Using recursion on left subtree
• Traversing both subtrees unnecessarily
• Returning root value with duplicates