BST Find Maximum Element in DSA Rust - Time & Space Complexity

Time Complexity: BST Find Maximum Element

O(h)

Understanding Time Complexity

We want to know how long it takes to find the largest value in a Binary Search Tree (BST).

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn find_max(root: &Option>) -> Option {
    let mut current = root;
    while let Some(node) = current {
        if node.right.is_none() {
            return Some(node.val);
        }
        current = &node.right;
    }
    None
}
    

This code finds the maximum value by moving right until no more right child exists.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving from one node to its right child in a loop.
• How many times: At most once per level down the right side of the tree.

How Execution Grows With Input

As the tree grows, the number of steps depends on the height of the rightmost path.

Input Size (n)	Approx. Operations
10	Up to 10 steps if tree is skewed right
100	Up to 100 steps if skewed right
1000	Up to 1000 steps if skewed right

Pattern observation: The time grows linearly with the height of the rightmost path, which can be as large as the number of nodes in worst case.

Final Time Complexity

Time Complexity: O(h)

This means the time depends on the height of the tree, moving down the right side until the maximum is found.

Common Mistake

[X] Wrong: "Finding the max is always O(1) because we just look at the root."

[OK] Correct: The max is at the rightmost node, not always the root, so we must follow right children which can take time proportional to tree height.

Interview Connect

Understanding how tree height affects search time helps you explain and optimize tree operations clearly in interviews.

Self-Check

What if the BST was balanced? How would the time complexity change?

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