# How to Validate if a Given Binary Tree is a Valid Binary Search Tree (BST)

> Learn how to validate if a binary tree is a valid Binary Search Tree BST. Discover the depth-first traversal method with permissible value ranges to ensure BST integrity.

- Repository: [Kevin Naughton Jr./interviews](https://github.com/kdn251/interviews)
- Tags: how-to-guide
- Published: 2026-03-04

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**You can validate a binary search tree by performing a depth-first traversal while propagating permissible value ranges (min, max) to each node, ensuring every value falls strictly within its assigned bounds.**

The `kdn251/interviews` repository provides multiple Java implementations demonstrating how to validate if a given binary tree adheres to BST ordering rules. This article examines the range-based validation algorithm used throughout the codebase, including two distinct approaches for handling boundary conditions in [`ValidBinarySearchTree.java`](https://github.com/kdn251/interviews/blob/main/ValidBinarySearchTree.java) and [`ValidateBinarySearchTree.java`](https://github.com/kdn251/interviews/blob/main/ValidateBinarySearchTree.java).

## Understanding the BST Validation Problem

A binary search tree (BST) must satisfy the **ordering invariant** for every node: all values in the left subtree must be strictly less than the node's value, and all values in the right subtree must be strictly greater. Merely checking that a node's immediate children obey this rule is insufficient—descendant nodes anywhere in the subtree must also satisfy the constraint relative to the original ancestor.

For example, a node with value 5 might have a right child with value 10, but if that right child has a left child with value 3, the tree is invalid because 3 violates the lower bound established by the root.

## The Range-Based Validation Algorithm

The most reliable method to verify BST properties is a **depth-first traversal** that tracks the permissible value range for each node. Instead of comparing only against the parent, each node inherits constraints from all its ancestors.

The algorithm works as follows:

1. **Initialize** the recursion with the full integer range (negative infinity to positive infinity).
2. **At each node**, verify the value lies strictly within the open interval `(min, max)`. If not, return `false`.
3. **Recurse** on the left child with range `(min, node.val)` and the right child with range `(node.val, max)`.
4. **Base case**: A `null` child is always valid.

This approach guarantees that every node satisfies the BST constraint relative to all its ancestors, not just its immediate parent.

## Java Implementation Examples

The `kdn251/interviews` repository contains two concrete implementations of this algorithm, each handling boundary conditions differently.

### Using Integer Bounds with Null Checks

The file [`cracking-the-coding-interview/chapter-four-trees-and-graphs/ValidBinarySearchTree.java`](https://github.com/kdn251/interviews/blob/main/cracking-the-coding-interview/chapter-four-trees-and-graphs/ValidBinarySearchTree.java) implements the range check using `Integer` objects, where `null` represents an unbounded side. The core logic resides in the `checkBST` method on lines 8-19.

```java
public class ValidBinarySearchTree {
    boolean checkBST(TreeNode n) {
        return checkBST(n, null, null);
    }
    
    boolean checkBST(TreeNode n, Integer min, Integer max) {
        if (n == null) return true;
        
        if ((min != null && n.val <= min) || (max != null && n.val >= max)) {
            return false;
        }
        
        return checkBST(n.left, min, n.val) && checkBST(n.right, n.val, max);
    }
}

```

This implementation requires null checks at each comparison but works correctly for standard integer ranges found in most interview scenarios.

### Using Long Sentinel Values

The file [`leetcode/tree/ValidateBinarySearchTree.java`](https://github.com/kdn251/interviews/blob/main/leetcode/tree/ValidateBinarySearchTree.java) (also found in [`company/facebook/ValidateBinarySearchTree.java`](https://github.com/kdn251/interviews/blob/main/company/facebook/ValidateBinarySearchTree.java)) avoids null checks by using `long` sentinel values for the initial bounds. The recursive method `validBSTRecursive` is defined on lines 37-44.

```java
public class ValidateBinarySearchTree {
    public boolean isValidBST(TreeNode root) {
        return validBSTRecursive(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    private boolean validBSTRecursive(TreeNode root, long min, long max) {
        if (root == null) return true;
        
        if (root.val <= min || root.val >= max) {
            return false;
        }
        
        return validBSTRecursive(root.left, min, root.val) && 
               validBSTRecursive(root.right, root.val, max);
    }
}

```

This approach is generally safer for extreme integer values because it uses the wider `long` range for initial bounds, preventing overflow issues when comparing against `Integer.MIN_VALUE` or `Integer.MAX_VALUE`.

### Example Usage

Both implementations validate the same tree structures:

```java
// Minimal TreeNode definition used by the repository
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

// Example 1 – a valid BST
TreeNode rootValid = new TreeNode(2);
rootValid.left  = new TreeNode(1);
rootValid.right = new TreeNode(3);

ValidateBinarySearchTree validator = new ValidateBinarySearchTree();
System.out.println(validator.isValidBST(rootValid)); // prints true

// Example 2 – an invalid BST (left child greater than root)
TreeNode rootInvalid = new TreeNode(2);
rootInvalid.left  = new TreeNode(3);   // violates BST rule
rootInvalid.right = new TreeNode(1);

System.out.println(validator.isValidBST(rootInvalid)); // prints false

```

## Complexity Analysis

Both implementations share the same computational characteristics:

- **Time Complexity**: **O(N)**, where *N* is the number of nodes in the tree. The algorithm visits each node exactly once to verify its value against the permissible range.
- **Space Complexity**: **O(H)**, where *H* is the height of the tree. This accounts for the recursion stack depth. In the worst case (a completely unbalanced tree), *H* equals *N*, resulting in **O(N)** space. For a balanced tree, the space complexity is **O(log N)**.

## Summary

- **Validate if a given binary tree is a valid Binary Search Tree** by performing a depth-first traversal while propagating valid value ranges (`min`, `max`) to each node.
- The `kdn251/interviews` repository provides two robust Java implementations: [`ValidBinarySearchTree.java`](https://github.com/kdn251/interviews/blob/main/ValidBinarySearchTree.java) using `Integer` bounds with `null` checks, and [`ValidateBinarySearchTree.java`](https://github.com/kdn251/interviews/blob/main/ValidateBinarySearchTree.java) using `long` sentinel values.
- Both approaches guarantee **O(N)** time complexity and correctly enforce the strict ordering invariant required for BST validation.

## Frequently Asked Questions

### What is the most efficient way to validate a BST?

The most efficient way to validate a BST is the **range-based depth-first search** algorithm, which achieves **O(N)** time complexity by visiting each node exactly once while tracking permissible value bounds. This approach is more efficient than in-order traversal with array storage, which requires **O(N)** additional space, and more reliable than simple parent-child comparisons that fail to catch ancestor constraint violations.

### Why use Long instead of Integer for BST validation?

Using `Long` (specifically `Long.MIN_VALUE` and `Long.MAX_VALUE`) as sentinel bounds prevents **integer overflow** issues when the tree contains values at the extreme ends of the `int` range. If you initialize bounds with `Integer.MIN_VALUE` or `Integer.MAX_VALUE`, a node with that exact value would fail the boundary check incorrectly. The `long` type provides a wider range that safely encompasses all possible `int` values.

### Can I validate a BST using in-order traversal?

Yes, you can validate a BST using **in-order traversal** because a valid BST produces a strictly increasing sequence when traversed in-order. The algorithm performs an in-order traversal while keeping track of the previously visited node's value; if the current node's value is not greater than the previous value, the tree is invalid. However, this approach typically requires **O(N)** space to store the traversal sequence or **O(H)** space for a recursive implementation with state tracking, making the range-based method more intuitive for most interview scenarios.

### What is the time complexity of BST validation?

The time complexity of BST validation is **O(N)**, where *N* represents the number of nodes in the tree, because any correct algorithm must examine each node at least once to verify the BST property. The auxiliary space complexity is **O(H)**, where *H* is the height of the tree, due to the recursion stack. In a balanced BST, this results in **O(log N)** space, while a completely unbalanced (skewed) tree requires **O(N)** space.