Why Validate if Tree is BST in DSA Rust?

The Big Idea

What if you could instantly know your tree is perfectly ordered without checking every node by hand?

The Scenario

Imagine you have a family tree drawn on paper. You want to check if the names are arranged in alphabetical order from left to right at every level. Doing this by eye is hard and mistakes happen easily.

The Problem

Manually checking each node and comparing it with others is slow and confusing. You might miss a mistake or get lost in the tree branches. It's easy to forget which nodes you already checked or to compare wrong pairs.

The Solution

Using a simple rule and a step-by-step check, you can quickly confirm if the tree follows the order rules of a Binary Search Tree (BST). This method looks at each node and ensures all left children are smaller and right children are bigger, making the check fast and reliable.

Before vs After

✗ Before

fn is_bst_manual(root: &Option<Box<TreeNode>>) -> bool {
    // Manually compare nodes without order rules
    // Confusing and error-prone
    true // placeholder
}

✓ After

fn is_bst(root: &Option<Box<TreeNode>>, min: Option<i32>, max: Option<i32>) -> bool {
    match root {
        None => true,
        Some(node) => {
            if let Some(min_val) = min {
                if node.val <= min_val { return false; }
            }
            if let Some(max_val) = max {
                if node.val >= max_val { return false; }
            }
            is_bst(&node.left, min, Some(node.val)) &&
            is_bst(&node.right, Some(node.val), max)
        }
    }
}

What It Enables

This check lets you trust that your tree data is correctly ordered, enabling fast searching and reliable data handling.

Real Life Example

When building a contact list app, ensuring the contacts are stored in a BST allows quick search by name, making the app fast and user-friendly.

Key Takeaways

Manual checks are slow and error-prone.

BST validation uses simple rules to check order efficiently.

Validating BST ensures fast and reliable data operations.

Practice

1. What does it mean for a binary tree to be a Binary Search Tree (BST)?

easy

A. Every node's left subtree contains only nodes with values less than the node's value, and the right subtree only nodes with values greater.

B. Every node has exactly two children.

C. All nodes have the same value.

D. The tree is balanced with equal height on both sides.

Solution

1. Step 1: Understand BST property

   A BST requires that for each node, all values in the left subtree are less than the node's value, and all values in the right subtree are greater.

2. Step 2: Compare options with definition

   Every node's left subtree contains only nodes with values less than the node's value, and the right subtree only nodes with values greater. correctly states this property, while others describe different tree properties.

3. Final Answer:

   Every node's left subtree contains only nodes with values less than the node's value, and the right subtree only nodes with values greater. -> Option A

4. Quick Check:

   BST property = Every node's left subtree contains only nodes with values less than the node's value, and the right subtree only nodes with values greater. [OK]

Hint: BST means left < node < right for all nodes [OK]

Common Mistakes:

• Confusing BST with balanced tree
• Thinking all nodes must have two children
• Assuming all values are equal

2. Which Rust function signature correctly represents a function to check if a binary tree is a BST?

easy

A. fn is_bst(root: &TreeNode) -> i32

B. fn is_bst(root: TreeNode) -> bool

C. fn is_bst(root: &Option) -> bool

D. fn is_bst(root: Option<TreeNode>) -> String

Solution

1. Step 1: Understand Rust tree node usage

   Binary trees in Rust are often represented as Option to allow for empty nodes and heap allocation.

2. Step 2: Check function signature return type

   The function should return a boolean indicating if the tree is a BST, so return type bool is correct.

3. Final Answer:

   fn is_bst(root: &Option) -> bool -> Option C

4. Quick Check:

   Correct Rust signature = fn is_bst(root: &Option) -> bool [OK]

Hint: Use &Option and bool return for BST check [OK]

Common Mistakes:

• Using TreeNode directly without Option or Box
• Returning wrong type like i32 or String
• Not using reference for input parameter

3. Given the following Rust code snippet, what will be the output?

struct TreeNode {
    val: i32,
    left: Option,
    right: Option,
}

fn is_bst(node: &Option, min: Option<i32>, max: Option<i32>) -> bool {
    match node {
        None => true,
        Some(n) => {
            if let Some(min_val) = min {
                if n.val <= min_val { return false; }
            }
            if let Some(max_val) = max {
                if n.val >= max_val { return false; }
            }
            is_bst(&n.left, min, Some(n.val)) && is_bst(&n.right, Some(n.val), max)
        }
    }
}

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 10,
        left: Some(Box::new(TreeNode { val: 5, left: None, right: None })),
        right: Some(Box::new(TreeNode { val: 15, left: None, right: None })),
    }));
    println!("{}", is_bst(&root, None, None));
}

medium

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Analyze tree structure

   The root node is 10, left child 5, right child 15. This satisfies BST property: 5 < 10 < 15.

2. Step 2: Trace is_bst function calls

   Function checks node values against min and max bounds recursively. All nodes satisfy constraints, so returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Tree is BST = true [OK]

Hint: Check node values within min/max bounds recursively [OK]

Common Mistakes:

• Ignoring min/max bounds in recursion
• Confusing <= and < in comparisons
• Assuming tree is not BST without checking children

4. Identify the bug in this Rust function that checks if a tree is a BST:

fn is_bst(node: &Option, min: Option<i32>, max: Option<i32>) -> bool {
    match node {
        None => true,
        Some(n) => {
            if let Some(min_val) = min {
                if n.val < min_val { return false; }
            }
            if let Some(max_val) = max {
                if n.val > max_val { return false; }
            }
            is_bst(&n.left, min, Some(n.val)) && is_bst(&n.right, Some(n.val), max)
        }
    }
}

medium

A. The function does not handle empty trees

B. The comparisons should use <= and >= instead of < and >

C. The recursion calls have wrong parameters order

D. The function should return i32 instead of bool

Solution

1. Step 1: Review comparison operators in BST validation

   BST requires node values to be strictly greater than min and strictly less than max, so using < and > is incorrect for duplicates.

2. Step 2: Correct comparison operators

   Comparisons should be <= min_val and >= max_val to reject equal values and maintain BST property.

3. Final Answer:

   The comparisons should use <= and >= instead of < and > -> Option B

4. Quick Check:

   Use <= and >= for BST checks [OK]

Hint: Use <= and >= to reject equal values in BST check [OK]

Common Mistakes:

• Using < and > allows duplicates breaking BST
• Mixing up parameter order in recursion
• Ignoring None case for empty trees

5. You have a binary tree where some nodes have duplicate values equal to their parent. You want to validate if it is a BST allowing duplicates only on the right subtree. How should you modify the Rust BST validation function?

hard

A. Disallow all duplicates by using strict < and > comparisons

B. Allow equal values only on the right by using n.val <= min_val for right subtree and n.val < max_val for left subtree

C. Allow equal values only when moving to the right subtree by changing comparison to n.val < min_val and n.val > max_val

D. Allow equal values only on the right by using n.val >= min_val for right subtree and n.val < max_val for left subtree

Solution

1. Step 1: Understand BST with duplicates on right

   Duplicates allowed only on right means left subtree nodes must be strictly less than current node, right subtree nodes can be equal or greater.

2. Step 2: Adjust min and max comparisons accordingly

   For left subtree, use n.val < max_val (strict less), for right subtree use n.val >= min_val (allow equal or greater).

3. Final Answer:

   Allow equal values only on the right by using n.val >= min_val for right subtree and n.val < max_val for left subtree -> Option D

4. Quick Check:

   Duplicates right subtree = n.val >= min_val [OK]

Hint: Allow equals only on right: use >= min_val, < max_val [OK]

Common Mistakes:

• Allowing duplicates on left subtree
• Using <= for left subtree allowing duplicates
• Not adjusting comparisons for duplicates