Why BST Find Maximum Element in DSA Rust?

The Big Idea

What if you could find the biggest number instantly without looking at every single one?

The Scenario

Imagine you have a messy pile of books with different heights, and you want to find the tallest book. If you check each book one by one, it takes a lot of time and effort.

The Problem

Manually checking each book is slow and easy to make mistakes, especially if the pile is huge. You might lose track or miss the tallest book.

The Solution

A Binary Search Tree (BST) organizes books so that the tallest book is always on the right side. To find the tallest book, you just keep going right until you can't go further. This is fast and simple.

Before vs After

✗ Before

let mut max_height = books[0].height;
for book in books.iter() {
    if book.height > max_height {
        max_height = book.height;
    }
}

✓ After

fn find_max(node: &Option<Box<Node>>) -> Option<i32> {
    match node {
        None => None,
        Some(n) => {
            if n.right.is_none() {
                Some(n.value)
            } else {
                find_max(&n.right)
            }
        }
    }
}

What It Enables

This lets you quickly find the largest value in a sorted structure without checking every element.

Real Life Example

Finding the highest score in a game leaderboard stored as a BST, so you instantly know the top player.

Key Takeaways

Manual search is slow and error-prone.

BST structure keeps largest elements on the right.

Finding max in BST is fast by going right until no more right child.

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