Validate if Tree is BST in DSA Rust - Time & Space Complexity

Time Complexity: Validate if Tree is BST

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a tree follows the rules of a Binary Search Tree (BST).

The question is: how does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn is_bst(node: &Option>, min: Option, max: Option) -> 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)
        }
    }
}

This code checks if a binary tree is a BST by visiting each node once and verifying its value fits the allowed range.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive visit of each node in the tree.
• How many times: Once per node, so total visits equal the number of nodes (n).

How Execution Grows With Input

As the tree grows, the function checks each node once, so the work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 checks
100	About 100 checks
1000	About 1000 checks

Pattern observation: The time grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check the tree grows directly in proportion to the number of nodes.

Common Mistake

[X] Wrong: "Checking if a tree is a BST only needs to look at the root or a few nodes."

[OK] Correct: Every node must be checked to ensure it fits the BST rules, so skipping nodes can miss violations.

Interview Connect

Understanding this time complexity helps you explain how efficient your tree checks are, a common skill in coding interviews.

Self-Check

"What if we used an in-order traversal to check the BST property instead? How would the time complexity change?"

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