Validate if Tree is BST in DSA C++ - Build from Scratch

Validate if Tree is BST

📖 Scenario: You are working with a simple binary tree structure in C++. You want to check if this tree follows the rules of a Binary Search Tree (BST). A BST is a tree where for every node, all nodes in the left subtree have smaller values, and all nodes in the right subtree have larger values.

🎯 Goal: Build a program that creates a binary tree, sets up a helper function to check BST rules, and then prints whether the tree is a valid BST or not.

📋 What You'll Learn

Create a binary tree with exactly 3 nodes with values 10, 5, and 15

Add a helper function to check if the tree is a BST using recursion

Use the helper function to determine if the tree is a BST

Print "BST" if the tree is valid, otherwise print "Not BST"

💡 Why This Matters

🌍 Real World

Checking if a tree is a BST is important in databases and search algorithms where data needs to be organized for fast lookup.

💼 Career

Understanding BST validation helps in roles involving data structures, algorithm design, and software engineering where tree data is common.

1

Create the binary tree nodes

Create a struct called Node with an int data and two pointers left and right. Then create three nodes named root, leftChild, and rightChild with values 10, 5, and 15 respectively. Connect leftChild as the left child of root and rightChild as the right child of root.

DSA C++

// Define Node struct and create root, leftChild, rightChild nodes
// Connect leftChild and rightChild to root
// Your code here

Hint

Define the Node struct first. Then create nodes using new Node{value, nullptr, nullptr}. Finally, link children to root.

2

Add helper function to check BST

Add a function called isBST that takes a Node* called node, and two int parameters min and max. This function returns true if the subtree rooted at node is a BST with all values strictly between min and max. If node is nullptr, return true. Otherwise, check if node->data is within the range, and recursively check left and right subtrees with updated ranges.

DSA C++

#include 
using namespace std;

struct Node {
    int data;
    Node* left;
    Node* right;
};

// Add isBST function here

int main() {
    Node* root = new Node{10, nullptr, nullptr};
    Node* leftChild = new Node{5, nullptr, nullptr};
    Node* rightChild = new Node{15, nullptr, nullptr};
    root->left = leftChild;
    root->right = rightChild;
    // Your code here
    return 0;
}

Hint

Use recursion to check left subtree with max as current node's data, and right subtree with min as current node's data.

3

Use the helper function to check the tree

In main, call isBST with root, INT_MIN, and INT_MAX as arguments. Store the result in a boolean variable called result.

DSA C++

#include 
#include 
using namespace std;

struct Node {
    int data;
    Node* left;
    Node* right;
};

bool isBST(Node* node, int min, int max) {
    if (node == nullptr) {
        return true;
    }
    if (node->data data >= max) {
        return false;
    }
    return isBST(node->left, min, node->data) && isBST(node->right, node->data, max);
}

int main() {
    Node* root = new Node{10, nullptr, nullptr};
    Node* leftChild = new Node{5, nullptr, nullptr};
    Node* rightChild = new Node{15, nullptr, nullptr};
    root->left = leftChild;
    root->right = rightChild;
    bool result = false;
    // Your code here
    return 0;
}

Hint

Call isBST with root, INT_MIN, and INT_MAX. Store the result in result.

4

Print if the tree is BST or not

Add a print statement that outputs "BST" if result is true, otherwise outputs "Not BST".

DSA C++

#include 
#include 
using namespace std;

struct Node {
    int data;
    Node* left;
    Node* right;
};

bool isBST(Node* node, int min, int max) {
    if (node == nullptr) {
        return true;
    }
    if (node->data data >= max) {
        return false;
    }
    return isBST(node->left, min, node->data) && isBST(node->right, node->data, max);
}

int main() {
    Node* root = new Node{10, nullptr, nullptr};
    Node* leftChild = new Node{5, nullptr, nullptr};
    Node* rightChild = new Node{15, nullptr, nullptr};
    root->left = leftChild;
    root->right = rightChild;
    bool result = isBST(root, INT_MIN, INT_MAX);
    if (result) {
        // print BST
    } else {
        // print Not BST
    }
    return 0;
}

Hint

Use cout to print "BST" if result is true, else print "Not BST".

Practice

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

easy

A. Every node has exactly two children.

B. For every node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger.

C. All leaf nodes are at the same depth.

D. The tree is perfectly balanced.

Solution

1. Step 1: Understand BST property

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

2. Step 2: Compare other options

   Options B, C, and D describe other tree properties but not BST specifically.

3. Final Answer:

   For every node, all nodes in the left subtree are smaller and all nodes in the right subtree are larger. -> Option B

4. Quick Check:

   BST property = left < node < right [OK]

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

Common Mistakes:

• Confusing BST with balanced tree
• Thinking all nodes must have two children
• Assuming leaf nodes must be at same depth

2. Which of the following C++ function signatures is correct to check if a binary tree is a BST using min and max limits?

easy

A. bool isBST(Node* root, int min, int max);

B. bool isBST(Node root, int min, int max);

C. bool isBST(Node* root, int min);

D. bool isBST(Node root);

Solution

1. Step 1: Check parameter types

   To check BST, we need a pointer to the root node and min/max limits to validate node values.

2. Step 2: Validate function signature

   bool isBST(Node* root, int min, int max); uses Node* root and both min and max parameters, which is correct for BST validation.

3. Final Answer:

   bool isBST(Node* root, int min, int max); -> Option A

4. Quick Check:

   Use pointer and min/max for BST check [OK]

Hint: Use pointer and min/max parameters for BST validation [OK]

Common Mistakes:

• Passing Node by value instead of pointer
• Missing max parameter
• Using incorrect parameter types

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

#include <iostream>
#include <climits>

struct Node {
  int data;
  Node* left;
  Node* right;
};

bool isBSTUtil(Node* root, int min, int max) {
  if (!root) return true;
  if (root->data <= min || root->data >= max) return false;
  return isBSTUtil(root->left, min, root->data) && isBSTUtil(root->right, root->data, max);
}

bool isBST(Node* root) {
  return isBSTUtil(root, INT_MIN, INT_MAX);
}

int main() {
  Node n1 = {10, nullptr, nullptr};
  Node n2 = {5, nullptr, nullptr};
  Node n3 = {15, nullptr, nullptr};
  n1.left = &n2;
  n1.right = &n3;
  std::cout << (isBST(&n1) ? "Yes" : "No") << std::endl;
  return 0;
}

medium

A. Runtime error

B. No

C. Compilation error

D. Yes

Solution

1. Step 1: Analyze tree structure

   Root is 10, left child 5 < 10, right child 15 > 10, so BST property holds.

2. Step 2: Check BST validation function

   Function correctly checks node values within min and max limits recursively.

3. Final Answer:

   Yes -> Option D

4. Quick Check:

   Tree satisfies BST rules = Yes [OK]

Hint: Check left < root < right for all nodes [OK]

Common Mistakes:

• Confusing <= and < in comparisons
• Not passing correct min/max values
• Ignoring nullptr base case

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

bool isBSTUtil(Node* root, int min, int max) {
  if (!root) return true;
  if (root->data < min || root->data > max) return false;
  return isBSTUtil(root->left, min, root->data) && isBSTUtil(root->right, root->data, max);
}

medium

A. The function should not use recursion.

B. The base case should return false when root is nullptr.

C. The comparison should use <= min and >= max to exclude equal values.

D. The min and max parameters should be swapped in recursive calls.

Solution

1. Step 1: Check comparison operators

   BST requires strict inequality: node values must be greater than min and less than max, so use <= and >= to reject equal values.

2. Step 2: Validate base case and recursion

   Base case returning true for nullptr is correct; recursion is necessary; min and max parameters are correctly passed.

3. Final Answer:

   The comparison should use <= min and >= max to exclude equal values. -> Option C

4. Quick Check:

   Use strict inequalities to validate BST [OK]

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

Common Mistakes:

• Allowing equal values breaks BST property
• Returning false on nullptr causes errors
• Swapping min and max in recursion

5. Given a binary tree where nodes can have duplicate values, how would you modify the BST validation function to allow duplicates only on the right subtree?

hard

A. Allow duplicates on right subtree by changing left subtree check to root->data < min and right subtree check to root->data >= max.

B. Change the left subtree check to root->data <= min and right subtree check to root->data > max.

C. Allow duplicates on left subtree by changing comparison to root->data <= min.

D. Disallow duplicates completely by using strict inequalities everywhere.

Solution

1. Step 1: Understand duplicate placement rule

   Duplicates allowed only on right subtree means left subtree nodes must be strictly less than root.

2. Step 2: Adjust comparison operators

   Left subtree check uses < min (strict), right subtree check uses >= max (allow equal or greater).

3. Final Answer:

   Allow duplicates on right subtree by changing left subtree check to root->data < min and right subtree check to root->data >= max. -> Option A

4. Quick Check:

   Duplicates only right = left < root, right >= root [OK]

Hint: Allow duplicates on right by using >= in right subtree check [OK]

Common Mistakes:

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