Validate if Tree is BST in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does BST stand for and what is its key property?

BST stands for Binary Search Tree. Its key property is that for every node, all nodes in its left subtree have smaller values, and all nodes in its right subtree have larger values.

beginner

What is the main approach to validate if a binary tree is a BST?

The main approach is to check if the tree satisfies the BST property at every node by ensuring all nodes in the left subtree are less than the node and all nodes in the right subtree are greater, often using min and max value limits during traversal.

intermediate

Why can't we just check if the left child is less and right child is greater to validate BST?

Because the BST property applies to all descendants, not just immediate children. A node deep in the left subtree must be less than the root, so checking only direct children is not enough.

intermediate

What role do min and max values play in BST validation?

Min and max values keep track of the allowed range for node values during recursion. Each node must be within these limits to satisfy BST rules.

beginner

What is the time complexity of validating a BST using the min-max approach?

The time complexity is O(n), where n is the number of nodes, because each node is visited once during the validation.

What must be true for every node in a BST?

A Node values are in ascending order from left to right

B Left child is always smaller than right child

C All nodes have two children

D All nodes in the left subtree are smaller and all nodes in the right subtree are larger

The BST property requires all nodes in the left subtree to be smaller and all nodes in the right subtree to be larger than the current node.

Which method helps track allowed values during BST validation?

A Using min and max value limits

B Checking only immediate children

C Counting nodes

D Sorting the tree

Min and max limits help ensure each node's value fits the BST property during recursive checks.

What is the time complexity of validating a BST with the min-max approach?

A O(log n)

B O(n)

C O(n^2)

D O(1)

Each node is visited once, so the time complexity is linear, O(n).

Why is checking only left and right child values not enough to validate a BST?

A Because trees can be unbalanced

B Because children can have equal values

C Because BST property applies to all descendants, not just children

D Because nodes can have more than two children

BST rules apply to all nodes in subtrees, so checking only immediate children misses violations deeper in the tree.

What does a valid BST output look like after validation?

A True if valid, False if not

B Sorted list of node values

C Height of the tree

D Number of nodes

Validation returns True if the tree satisfies BST properties, otherwise False.

Explain how to validate if a binary tree is a BST using min and max values.

Think about how allowed value ranges change as you go down the tree.

Describe why checking only immediate children is insufficient to validate a BST.

Consider a node deep in the left subtree that violates BST rules.

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