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]) return true;

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

A0

Bnullptr

CNULL

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or NULL instead of nullptr

Returning false instead of true here

We check if the node pointer is nullptr to know if we reached a leaf's child.

2fill in blank

medium

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

DSA C++

if (root->val <= [1] || root->val >= maxVal) return false;

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

Aroot->val

BmaxVal

CminVal

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxVal instead of minVal here

Comparing with root->val itself

We check if the node's value is less than or equal to minVal or greater than or equal to maxVal.

3fill in blank

hard

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

DSA C++

return isValidBST(root->left, [1], root->val) && isValidBST(root->right, root->val, maxVal);

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

AminVal

Bnullptr

Croot->val

DmaxVal

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxVal instead of minVal for left subtree

Passing nullptr instead of a value

The left subtree must have values less than the current node's value, so pass minVal as lower bound and update max to root->val.

4fill in blank

hard

Fill both blanks to complete the function signature and initial call for BST validation.

DSA C++

bool isValidBST(TreeNode* root, [1] minVal = [2], long maxVal = LONG_MAX) {

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

Along

Bint

Cnullptr

DLONG_MIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using int instead of long

Using nullptr as default for minVal

The function uses long type for minVal and maxVal to handle edge cases. The initial minVal is set to LONG_MIN.

5fill in blank

hard

Fill all three blanks to complete the recursive BST validation logic.

DSA C++

if (root == nullptr) return true;
if (root->val <= [1] || root->val >= [2]) return false;
return isValidBST(root->left, [1], root->val) && isValidBST(root->right, root->val, [2]);

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

AminVal

BmaxVal

Croot->val

Dnullptr

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up minVal and maxVal

Using root->val incorrectly in comparisons

The node's value must be greater than minVal and less than maxVal. The recursive calls update these bounds accordingly.

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