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

Validate if Tree is BST

📖 Scenario: Imagine you have a family tree stored in a computer program. You want to check if this tree follows the rules of a Binary Search Tree (BST), where each node's left child is smaller and the right child is larger. This helps in quickly finding family members.

🎯 Goal: You will write a program in C to create a simple binary tree, set up a helper function to check if it is a BST, and then print the result.

📋 What You'll Learn

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

Create a helper function to check if the tree is a BST

Use recursion and limits to validate the BST property

Print 1 if the tree is a BST, otherwise print 0

💡 Why This Matters

🌍 Real World

Checking if a tree is a BST is important in databases and search engines to ensure fast data retrieval.

💼 Career

Understanding BST validation helps in roles involving data structures, algorithms, and software development where tree data is common.

1

Create the binary tree structure and nodes

Define a struct called Node with int data, Node* left, and Node* right. Then create three nodes with values 10, 5, and 15. Connect the node with value 10 as root, its left child as node with value 5, and right child as node with value 15.

DSA C

// Define struct Node and create nodes with values 10, 5, 15
// Connect nodes to form the tree

Hint

Use a struct to define the tree node. Use a helper function to create nodes easily.

2

Add helper function prototype and initial call

Declare a function int isBST(Node* node, int min, int max) that will check if the tree is a BST. In main, prepare to call isBST with root, INT_MIN, and INT_MAX. Include limits.h for these constants.

DSA C

#include 

int isBST(Node* node, int min, int max);

int main() {
    Node* root = createNode(10);
    root->left = createNode(5);
    root->right = createNode(15);
    
    // Prepare to call isBST with root, INT_MIN, INT_MAX
    
    return 0;
}

Hint

Include limits.h to use INT_MIN and INT_MAX. Declare the function prototype before main.

3

Implement the isBST function

Write the function int isBST(Node* node, int min, int max) that returns 1 if the subtree rooted at node is a BST within the limits min and max, otherwise returns 0. Use recursion to check left and right subtrees with updated limits.

DSA C

int isBST(Node* node, int min, int max) {
    // Your code here
}

Hint

Check if current node is NULL (return 1). Then check if node's data is within limits. Recursively check left and right subtrees with updated limits.

4

Print the result of BST validation

In main, call isBST(root, INT_MIN, INT_MAX) and print the returned value using printf. This will print 1 if the tree is a BST, otherwise 0.

DSA C

int main() {
    Node* root = createNode(10);
    root->left = createNode(5);
    root->right = createNode(15);

    // Print the result of isBST
    
    return 0;
}

Hint

Use printf("%d\n", isBST(root, INT_MIN, INT_MAX)); to print the result.

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