Validate if Tree is BST in DSA C++ - Time & Space Complexity

Time Complexity: Validate if Tree is BST

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a tree follows the rules of a Binary Search Tree (BST).

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

bool isBSTUtil(Node* node, int min, int max) {
    if (node == nullptr) return true;
    if (node->data <= min || node->data >= max) return false;
    return isBSTUtil(node->left, min, node->data) &&
           isBSTUtil(node->right, node->data, max);
}

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

This code checks if a binary tree is a BST by making sure every node fits the BST rules within a range.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

As the tree grows, the function visits every node exactly once.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The time grows directly in proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check the tree grows linearly with the number of nodes.

Common Mistake

[X] Wrong: "Checking if a tree is BST takes more than linear time because of nested checks."

[OK] Correct: Each node is checked once with simple comparisons, so the total work is just one pass through all nodes.

Interview Connect

Understanding this helps you explain how to efficiently verify tree properties, a common skill in coding interviews.

Self-Check

"What if the tree is given as a sorted array and you want to check if it can form a BST? How would the time complexity change?"

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