Why Validate if Tree is BST in DSA C?

The Big Idea

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

The Scenario

Imagine you have a family tree drawn on paper, and you want to check if it 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 person and comparing them manually is tiring and confusing.

The Problem

Checking each node by hand means you might forget some comparisons or mix up the order. It takes a lot of time and mistakes happen easily, especially if the tree is big or messy.

The Solution

Using a program to check if the tree follows the rules of a Binary Search Tree (BST) makes this easy and fast. The program looks at each node and makes sure all left children are smaller and all right children are bigger, without missing any part.

Before vs After

✗ Before

int isBST(Node* root) {
  if (root == NULL) return 1;
  if (root->left && root->left->data > root->data) return 0;
  if (root->right && root->right->data < root->data) return 0;
  return isBST(root->left) && isBST(root->right);
}

✓ After

#include <limits.h>

int isBSTUtil(Node* node, int min, int max) {
  if (node == NULL) return 1;
  if (node->data <= min || node->data >= max) return 0;
  return isBSTUtil(node->left, min, node->data) && isBSTUtil(node->right, node->data, max);
}

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

What It Enables

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

Real Life Example

When building a phone contact list that needs to be searched quickly, ensuring the underlying tree is a BST helps find contacts fast without errors.

Key Takeaways

Manual checking is slow and error-prone.

Using min-max limits helps verify BST property correctly.

Validating BST ensures efficient data operations.

Practice

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

easy

A. Each node has exactly two children

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

C. All nodes have unique values

D. The tree is perfectly 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 matches the BST property exactly, while others describe different tree properties.

3. Final Answer:

   For each 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 = 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 for all nodes [OK]

Common Mistakes:

• Confusing BST with balanced tree
• Thinking all nodes must have two children
• Assuming unique values define BST

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

easy

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

B. void isBST(struct Node* root);

C. int isBST(struct Node* root);

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

Solution

1. Step 1: Identify typical BST check function parameters

   Checking BST usually requires passing current node and min/max limits to validate subtree values.

2. Step 2: Match function signatures

   int isBST(struct Node* root, int min, int max); uses int return type and min/max parameters, which is common in C. bool isBST(struct Node* root, int min, int max); uses bool which is not standard C. Options B and C lack min/max parameters.

3. Final Answer:

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

4. Quick Check:

   BST check needs min/max limits and int return [OK]

Hint: Use min/max limits in function parameters for BST check [OK]

Common Mistakes:

• Omitting min/max parameters
• Using bool type in plain C
• Using void return type for boolean check

3. Given the following tree structure and code snippet, what will be the output?

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

int isBSTUtil(struct Node* node, int min, int max) {
  if (node == NULL) return 1;
  if (node->data <= min || node->data >= max) return 0;
  return isBSTUtil(node->left, min, node->data) && isBSTUtil(node->right, node->data, max);
}

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

// Tree:
//      10
//     /  \
//    5   15
//       /  \
//      6   20

int main() {
  struct Node n1 = {10, NULL, NULL};
  struct Node n2 = {5, NULL, NULL};
  struct Node n3 = {15, NULL, NULL};
  struct Node n4 = {6, NULL, NULL};
  struct Node n5 = {20, NULL, NULL};
  n1.left = &n2; n1.right = &n3;
  n3.left = &n4; n3.right = &n5;
  printf("%d\n", isBST(&n1));
  return 0;
}

medium

A. 1

B. Undefined behavior

C. Compilation error

D. 0

Solution

1. Step 1: Analyze tree structure and BST property

   The node with value 6 is in the right subtree of 10 but less than 10, violating BST property.

2. Step 2: Trace isBSTUtil calls

   When checking node 6, min=15 and max=INT_MAX, 6 <= 15 fails the condition, so returns 0.

3. Final Answer:

   0 -> Option D

4. Quick Check:

   Node 6 violates BST rules, so output = 0 [OK]

Hint: Check if any node violates min/max limits during recursion [OK]

Common Mistakes:

• Ignoring subtree min/max constraints
• Assuming all left < root and right > root only locally
• Not tracing recursive calls carefully

4. Identify the bug in this BST validation function and select the fix:

int isBST(struct Node* root) {
  if (root == NULL) return 1;
  if (root->left != NULL && root->left->data >= root->data) return 0;
  if (root->right != NULL && root->right->data <= root->data) return 0;
  if (!isBST(root->left) || !isBST(root->right)) return 0;
  return 1;
}

medium

A. Add min and max parameters to check entire subtree ranges

B. Change >= to > and <= to < in comparisons

C. Remove recursive calls to isBST

D. Return 0 if root is NULL

Solution

1. Step 1: Understand current checks

   The function only compares immediate children with root, missing deeper subtree violations.

2. Step 2: Explain why min/max parameters are needed

   To validate BST fully, we must ensure all nodes in left subtree < root and right subtree > root using min/max limits recursively.

3. Final Answer:

   Add min and max parameters to check entire subtree ranges -> Option A

4. Quick Check:

   Min/max limits fix subtree validation [OK]

Hint: Use min/max limits to validate entire subtree, not just immediate children [OK]

Common Mistakes:

• Checking only immediate children
• Ignoring subtree value ranges
• Misusing comparison operators

5. You want to validate if a binary tree is a BST but the tree may contain duplicate values allowed only on the right subtree. How would you modify the BST validation logic in C?

hard

A. Use strict > and < comparisons for both subtrees

B. Allow duplicates only on left subtree by changing conditions

C. Change the condition to allow node->data >= min and < max for right subtree

D. Ignore duplicates and check only structure

Solution

1. Step 1: Understand duplicate allowance rule

   Duplicates allowed only on right subtree means left subtree nodes must be strictly less, right subtree nodes can be equal or greater.

2. Step 2: Modify min/max checks accordingly

   For left subtree, use node->data > min and < root->data; for right subtree, allow node->data >= root->data and < max.

3. Final Answer:

   Change the condition to allow node->data >= min and < max for right subtree -> Option C

4. Quick Check:

   Allow duplicates on right by using >= on right subtree [OK]

Hint: Allow duplicates on right by using >= comparison there [OK]

Common Mistakes:

• Allowing duplicates on left subtree incorrectly
• Using strict inequalities everywhere
• Ignoring duplicate rules