Validate if Tree is BST in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is NULL.

DSA C

if (root [1] NULL) return 1;

Drag options to blanks, or click blank then click option'

A>

B!=

C<

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causes wrong logic.

Using comparison operators like '>' or '<' with pointers is invalid.

We check if the node is NULL using '=='. If it is NULL, it means the subtree is empty and thus a valid BST.

2fill in blank

medium

Complete the code to check if the current node's value is within the allowed range.

DSA C

if (root->data [1] min || root->data [2] max) return 0;

Drag options to blanks, or click blank then click option'

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' or '>' instead of '<=' or '>=' causes boundary errors.

Checking only one boundary instead of both.

The node's data must be greater than min and less than max. So we check if it is less than or equal to min or greater than or equal to max to return false.

3fill in blank

hard

Fix the error in the recursive call to validate the left subtree.

DSA C

return isBST(root->left, min, [1]) && isBST(root->right, root->data, max);

Drag options to blanks, or click blank then click option'

Aroot->data

Bmin

Cmax

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Passing min instead of root->data as max boundary.

Passing max instead of root->data.

For the left subtree, the maximum allowed value is the current node's data, so we pass root->data as the new max.

4fill in blank

hard

Fill both blanks to complete the recursive call for the right subtree and the initial call to isBST.

DSA C

return isBST(root->left, min, [1]) && isBST(root->right, [2], max);

Drag options to blanks, or click blank then click option'

Aroot->data

Bmin

Croot->data + 1

Dmax

Attempts:

3 left

💡 Hint

Common Mistakes

Using max for right subtree min boundary.

Using min for left subtree max boundary.

For the left subtree, max is root->data. For the right subtree, min is root->data. This keeps values strictly less or greater than current node.

5fill in blank

hard

Fill the three blanks to complete the isBST function signature. Note that the initial call uses INT_MIN and INT_MAX for the full integer range.

DSA C

int isBST(struct Node* [1], int [2], int [3]) {
    if (root == NULL) return 1;
    if (root->data <= min || root->data >= max) return 0;
    return isBST(root->left, min, root->data) && isBST(root->right, root->data, max);
}

// Initial call
int result = isBST(root, INT_MIN, INT_MAX);

Drag options to blanks, or click blank then click option'

Aroot

Bmin

Cmax

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Mismatching parameter names causing errors.

Using wrong variable names in initial call.

The function takes a pointer named root and two integers min and max. The initial call uses the full integer range as min and max.

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