BST Find Maximum Element in DSA Rust - Execution Trace

Concept Flow - BST Find Maximum Element

Start at root node

↓

Check if right child exists?

No→Current node is max

Yes↓

Move to right child

↩Back to check right child

Start at the root and keep moving to the right child until there is no right child. The last node reached is the maximum.

Execution Sample

DSA Rust

fn find_max(root: &Option<Box<Node>>) -> Option<i32> {
    let mut current = root;
    while let Some(node) = current {
        if node.right.is_none() {
            return Some(node.value);
        }
        current = &node.right;
    }
    None
}

This code finds the maximum value in a BST by moving right until no right child exists.

Execution Table

Step	Operation	Current Node Value	Right Child Exists?	Pointer Changes	Visual State
1	Start at root	15	Yes	current -> Node(15)	15 -> 10 -> 20 -> 8 -> 12 -> 17 -> 25
2	Check right child	15	Yes	current -> Node(20)	15 -> 10 -> 20 -> 8 -> 12 -> 17 -> 25
3	Check right child	20	Yes	current -> Node(25)	15 -> 10 -> 20 -> 8 -> 12 -> 17 -> 25
4	Check right child	25	No	current stays at Node(25)	15 -> 10 -> 20 -> 8 -> 12 -> 17 -> 25
5	Return max value	25	No	Return 25	15 -> 10 -> 20 -> 8 -> 12 -> 17 -> 25

💡 Right child of node 25 is None, so 25 is maximum.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
current	Node(15)	Node(20)	Node(25)	Node(25)	Node(25)

Key Moments - 3 Insights

Why do we keep moving to the right child to find the maximum?

Because in a BST, all values to the right are greater. The execution_table rows 2 and 3 show moving right to larger values.

What if the tree is empty (root is None)?

The code returns None immediately as current is None from the start, so no maximum exists.

Why do we stop when right child is None, not when current is None?

Stopping at right child None means current node is the largest. Execution_table row 4 shows this stopping condition.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'current' after Step 3?

A Node(20)

B Node(15)

C Node(25)

D None

💡 Hint

Check the 'Pointer Changes' column in row 3 of execution_table.

At which step does the condition 'Right Child Exists?' become No?

A Step 4

B Step 3

C Step 2

D Step 5

💡 Hint

Look at the 'Right Child Exists?' column in execution_table rows.

If the root node had no right child, what would be the maximum value returned?

A Value of left child

B Value of root node

C None

D Value of right child

💡 Hint

Refer to the concept_flow and execution_table step 1 where right child check is done.

Concept Snapshot

BST Find Maximum Element:
- Start at root node
- Move right while right child exists
- When no right child, current node is max
- Return current node's value
- If tree empty, return None

Full Transcript

To find the maximum element in a Binary Search Tree (BST), start at the root node. Check if the current node has a right child. If yes, move to that right child and repeat. Continue moving right until you find a node with no right child. That node contains the maximum value in the BST. If the tree is empty, return None. This process works because in BSTs, right children always hold greater values than their parents.

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