Validate if Tree is BST in DSA C

Mental Model

A binary search tree (BST) has nodes arranged so that left children are smaller and right children are larger than the parent. We check this by making sure every node fits in a valid range.

Analogy: Imagine a family tree where every child must be younger than their parent on the left side and older on the right side. We check each person's age fits this rule to confirm the tree is correct.

      10
     /  \
    5    15
   / \     \
  2   7     20

Dry Run Walkthrough

Input: Tree: 10 -> left:5, right:15; 5 -> left:2, right:7; 15 -> right:20

Goal: Check if this tree follows BST rules where left < node < right for all nodes

Step 1: Start at root 10 with allowed range (-∞, +∞)

10 [range: -∞, +∞]
 /  \
5    15

Why: We begin with no limits on root value

Step 2: Check left child 5 with range (-∞, 10)

5 [range: -∞, 10]
 / \
2   7

Why: Left child must be less than 10

Step 3: Check left child 2 with range (-∞, 5)

2 [range: -∞, 5]

Why: 2 must be less than 5

Step 4: Check right child 7 with range (5, 10)

7 [range: 5, 10]

Why: 7 must be greater than 5 and less than 10

Step 5: Check right child 15 with range (10, +∞)

15 [range: 10, +∞]
  \
  20

Why: Right child must be greater than 10

Step 6: Check right child 20 with range (15, +∞)

20 [range: 15, +∞]

Why: 20 must be greater than 15

Result:

Tree is a valid BST because all nodes fit their allowed ranges

Annotated Code

DSA C

#include <stdio.h>
#include <limits.h>
#include <stdbool.h>
#include <stdlib.h>

// Define tree node
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};

// Create new node
struct Node* newNode(int data) {
    struct Node* node = (struct Node*)malloc(sizeof(struct Node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
    return node;
}

// Recursive function to validate BST
bool isBSTUtil(struct Node* node, int min, int max) {
    if (node == NULL) return true; // empty tree is BST

    // Check current node data against allowed range
    if (node->data <= min || node->data >= max) return false;

    // Check left subtree with updated max
    if (!isBSTUtil(node->left, min, node->data)) return false;

    // Check right subtree with updated min
    if (!isBSTUtil(node->right, node->data, max)) return false;

    return true;
}

// Wrapper function
bool isBST(struct Node* root) {
    return isBSTUtil(root, INT_MIN, INT_MAX);
}

int main() {
    // Build tree from dry run example
    struct Node* root = newNode(10);
    root->left = newNode(5);
    root->right = newNode(15);
    root->left->left = newNode(2);
    root->left->right = newNode(7);
    root->right->right = newNode(20);

    if (isBST(root)) {
        printf("Tree is a valid BST\n");
    } else {
        printf("Tree is NOT a valid BST\n");
    }
    return 0;
}

if (node == NULL) return true;

empty subtree is valid BST base case

if (node->data <= min || node->data >= max) return false;

check current node fits allowed min/max range

if (!isBSTUtil(node->left, min, node->data)) return false;

validate left subtree with updated max limit

if (!isBSTUtil(node->right, node->data, max)) return false;

validate right subtree with updated min limit

Output

Tree is a valid BST

Complexity Analysis

Time: O(n) because we visit each node once to check its value

Space: O(h) where h is tree height due to recursion stack

vs Alternative: Naive approach checks all nodes repeatedly leading to O(n^2), this method uses range limits to check in one pass

Edge Cases

Empty tree

Returns true because empty tree is a valid BST

DSA C

if (node == NULL) return true;

Single node tree

Returns true because one node always satisfies BST rules

DSA C

if (node == NULL) return true;

Tree with duplicate values

Returns false because duplicates violate strict less/greater rule

DSA C

if (node->data <= min || node->data >= max) return false;

When to Use This Pattern

When you need to confirm if a binary tree follows the BST property, use range checks on each node recursively to validate the entire tree efficiently.

Common Mistakes

Mistake: Only checking immediate children without considering the full allowed range
Fix: Pass down min and max limits to each recursive call to ensure all descendants fit BST rules

Summary

Checks if a binary tree is a valid binary search tree by verifying node values fit allowed ranges.

Use when you need to confirm the BST property for a given binary tree structure.

The key insight is to track and update min and max limits for each node to ensure global BST validity.

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