Why Validate if Tree is BST in DSA Java?

The Big Idea

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

The Scenario

Imagine you have a family tree drawn on paper. You want to check if the tree follows a special rule: every child on the left is smaller than the parent, and every child on the right is bigger. Doing this by looking at each node and comparing manually is like checking every family member's age one by one without any order.

The Problem

Manually checking each node against all others is slow and confusing. You might miss some nodes or compare the wrong ones. It's easy to make mistakes and waste time, especially if the tree is big.

The Solution

Using a method to validate if the tree is a Binary Search Tree (BST) helps you quickly check if the tree follows the rules. It uses a smart way to compare nodes by keeping track of allowed value ranges as it moves through the tree, making the check fast and reliable.

Before vs After

✗ Before

boolean isBST(Node root) {
  // Check each node against all others manually
  // Very complex and slow
  return false;
}

✓ After

boolean isBST(Node node, int min, int max) {
  if (node == null) return true;
  if (node.value <= min || node.value >= max) return false;
  return isBST(node.left, min, node.value) && isBST(node.right, node.value, max);
}

What It Enables

This lets you quickly and correctly confirm if a tree is a BST, enabling efficient searching and sorting operations.

Real Life Example

When building a phone contact list app, ensuring the contact names are stored in a BST allows fast searching and adding new contacts without errors.

Key Takeaways

Manual checking is slow and error-prone.

Validating BST uses value limits to check nodes efficiently.

This method ensures fast and correct tree validation.

Practice

1. What is the main property that defines a Binary Search Tree (BST)?

easy

A. For each node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger

B. Each node has exactly two children

C. All nodes have unique values but no order is required

D. The tree is always balanced

Solution

1. Step 1: Understand BST property

   A BST requires that for every node, values in the left subtree are smaller and values in the right subtree are larger.

2. Step 2: Compare options with BST definition

   For each node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger correctly states this property, while others describe different tree characteristics.

3. Final Answer:

   For each node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger -> Option A

4. Quick Check:

   BST property = For each node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger [OK]

Hint: Remember BST means left < node < right [OK]

Common Mistakes:

• Confusing BST with balanced tree
• Thinking all nodes must have two children
• Ignoring the ordering property

2. Which of the following Java method signatures is correct to check if a binary tree is a BST?

easy

A. public int isBST(Node root, int min, int max)

B. public boolean isBST(Node root, int min, int max)

C. public void isBST(Node root)

D. public boolean isBST(Node root, Node min, Node max)

Solution

1. Step 1: Identify correct return type and parameters

   Checking BST returns true/false, so return type should be boolean. Also, min and max values are needed as integers to validate node values.

2. Step 2: Match method signature

   public boolean isBST(Node root, int min, int max) matches this: boolean return type and int min, max parameters. Others have wrong return types or parameter types.

3. Final Answer:

   public boolean isBST(Node root, int min, int max) -> Option B

4. Quick Check:

   Return boolean with int min/max params [OK]

Hint: Use boolean return and int min/max for range checks [OK]

Common Mistakes:

• Using void return type for a check method
• Using Node type for min/max instead of int
• Returning int instead of boolean

3. What will be the output of the following code snippet?

class Node {
  int val;
  Node left, right;
  Node(int v) { val = v; left = right = null; }
}

public class Test {
  static boolean isBST(Node root, int min, int max) {
    if (root == null) return true;
    if (root.val <= min || root.val >= max) return false;
    return isBST(root.left, min, root.val) && isBST(root.right, root.val, max);
  }

  public static void main(String[] args) {
    Node root = new Node(10);
    root.left = new Node(5);
    root.right = new Node(15);
    root.left.left = new Node(1);
    root.left.right = new Node(12);
    System.out.println(isBST(root, Integer.MIN_VALUE, Integer.MAX_VALUE));
  }
}

medium

A. Runtime exception

B. true

C. Compilation error

D. false

Solution

1. Step 1: Trace the tree nodes and BST conditions

   The node with value 12 is in the left subtree of 10 but greater than 10, violating BST property.

2. Step 2: Check isBST calls

   When checking root.left.right (12), it fails the condition root.val <= min or root.val >= max because 12 >= 10 is true, so returns false.

3. Final Answer:

   false -> Option D

4. Quick Check:

   Node 12 violates BST rule, so false [OK]

Hint: Check if any left child > root or right child < root [OK]

Common Mistakes:

• Ignoring subtree value ranges
• Assuming all nodes in left subtree are less than root only
• Not updating min/max during recursion

4. Identify the bug in this BST validation code snippet:

boolean isBST(Node root) {
  if (root == null) return true;
  if (root.left != null && root.left.val > root.val) return false;
  if (root.right != null && root.right.val < root.val) return false;
  return isBST(root.left) && isBST(root.right);
}

medium

A. It does not handle null root properly

B. It uses wrong comparison operators for left and right children

C. It does not check the entire subtree values, only immediate children

D. It should return int instead of boolean

Solution

1. Step 1: Analyze the checks performed

   The code only compares root with its immediate left and right children, not the entire subtree.

2. Step 2: Understand BST validation requirement

   BST validation requires all nodes in left subtree < root and all in right subtree > root, not just immediate children.

3. Final Answer:

   It does not check the entire subtree values, only immediate children -> Option C

4. Quick Check:

   Check full subtree ranges, not just children [OK]

Hint: Check full subtree ranges, not just direct children [OK]

Common Mistakes:

• Only comparing immediate children values
• Ignoring recursive min/max range checks
• Assuming local checks are enough

5. Given a binary tree, which approach correctly validates if it is a BST considering edge cases like duplicate values and integer limits?

hard

A. Use min and max integer range parameters updated during recursion

B. Use in-order traversal to check if values are strictly increasing

C. Check only immediate children values recursively

D. Compare each node with root only, ignoring subtrees

Solution

1. Step 1: Understand edge cases in BST validation

   Duplicates and integer limits require careful range checks, not just value comparisons.

2. Step 2: Evaluate approaches

   Using min and max parameters updated during recursion correctly enforces BST property for all nodes and handles edge cases.

3. Final Answer:

   Use min and max integer range parameters updated during recursion -> Option A

4. Quick Check:

   Range checks with min/max handle edge cases [OK]

Hint: Use min/max range checks recursively for full validation [OK]

Common Mistakes:

• Ignoring duplicates or equal values
• Not updating min/max during recursion
• Only checking immediate children