🚀Day 10. Check for BST 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given the root of a binary tree, check whether it is a Binary Search Tree (BST) or not.

Definition of BST:

• The left subtree of a node contains only nodes with keys less than the node's key.

• The right subtree of a node contains only nodes with keys greater than the node's key.

• Both the left and right subtrees must also be binary search trees.

• Note: BSTs cannot contain duplicate nodes.

🔍 Example Walkthrough:

Example 1:

Input:

root = [2, 1, 3, N, N, N, 5]

Output:

true

Explanation:

• The left subtree of every node contains smaller keys and right subtree of every node contains greater keys. Hence, it is a BST.

Example 2:

Input:

root = [2, N, 7, N, 6, N, 9]

Output:

false

Explanation:

• Since the node to the right of node with key 7 has lesser key value, Hence, it is not a BST.

Example 3:

Input:

root = [10, 5, 20, N, N, 9, 25]

Output:

false

Explanation:

• The node with key 9 present in the right subtree has lesser key value than root node. Hence, it is not a BST.

Constraints:

• 1 ≤ Number of nodes ≤ $10^5$

• 1 ≤ node->data ≤ $10^9$

🎯 My Approach:

✅ Min–Max Recursion (Top-Down Approach)

1. Base Case:

   • If the current node is NULL, return true (an empty tree is a valid BST).

2. Check Current Node:

   • The current node’s value should be greater than the min value and less than the max value.

3. Recursive Calls:

   • Recursively check the left subtree with the updated range [min, node->data].

   • Recursively check the right subtree with the updated range [node->data, max].

4. Return Result:

   • The tree is a BST only if both left and right subtrees are BSTs.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N) We visit each node exactly once, performing constant-time operations at each step.

• Expected Auxiliary Space Complexity: O(H) Due to the recursion stack, where H is the height of the tree. In the worst case (skewed tree), H = N. In the best case (balanced tree), H = log N.

📝 Solution Code

Code (C++)

class Solution {
public:
    bool isBST(Node* root, int min = INT_MIN, int max = INT_MAX) {
        return !root || (root->data > min && root->data < max &&
                         isBST(root->left, min, root->data) &&
                         isBST(root->right, root->data, max));
    }
};

🌲 Alternative Approaches

2️⃣ Inorder Traversal (Recursive)

• Perform an inorder traversal to produce a list of values.

• A BST’s inorder traversal should result in a strictly increasing sequence.

• If the sequence is not increasing, the tree is not a BST.

class Solution {
    void inorder(Node* root, vector<int>& vals) {
        if (!root) return;
        inorder(root->left, vals);
        vals.push_back(root->data);
        inorder(root->right, vals);
    }
public:
    bool isBST(Node* root) {
        vector<int> vals;
        inorder(root, vals);
        for (int i = 1; i < vals.size(); i++) {
            if (vals[i] <= vals[i-1]) return false;
        }
        return true;
    }
};

3️⃣ Iterative Inorder Traversal (Using Stack)

• Instead of recursion, use a stack to simulate inorder traversal.

• Compare each node’s value with the previous node’s value to check for the BST property.

class Solution {
public:
    bool isBST(Node* root) {
        stack<Node*> st;
        Node* prev = nullptr;
        while (root || !st.empty()) {
            while (root) {
                st.push(root);
                root = root->left;
            }
            root = st.top();
            st.pop();
            if (prev && root->data <= prev->data) return false;
            prev = root;
            root = root->right;
        }
        return true;
    }
};

📊 Comparison of Approaches

Approaches	⏱️ Time Complexity	🗂️ Space Complexity	⚡ Method	✅ Pros	⚠️ Cons
1️⃣ Min–Max Recursion	🟢 O(N)	🟡 O(H)	Recursive (Min–Max)	Fastest; no extra storage	Recursive depth may cause stack overflow
2️⃣ Inorder Traversal (Recursive)	🟢 O(N)	🟡 O(N)	Recursive Inorder	Simple implementation; easy to understand	Requires extra space for storing nodes
3️⃣ Iterative Inorder Traversal	🟢 O(N)	🟡 O(H)	Stack-based DFS	Avoids recursion stack overflow	Code complexity is slightly higher

💡 Best Choice?

• For Balanced Trees / Small Depth: ✅ Approach 1 (Min–Max Recursion) is the fastest and most efficient.

• For Deep or Skewed Trees (Risk of Stack Overflow): ✅ Approach 3 (Iterative Inorder Traversal) handles deep recursion better.

• For Simple Understanding & Learning: ✅ Approach 2 (Inorder Traversal Recursive) is the most intuitive to grasp.

Code (Java)

class Solution {
    boolean isBST(Node root) {
        return isBST(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }

    boolean isBST(Node node, int min, int max) {
        return node == null || (node.data > min && node.data < max &&
                isBST(node.left, min, node.data) &&
                isBST(node.right, node.data, max));
    }
}

Code (Python)

class Solution:
    def isBST(self, root, min_val=float('-inf'), max_val=float('inf')):
        return not root or (min_val < root.data < max_val and
                            self.isBST(root.left, min_val, root.data) and
                            self.isBST(root.right, root.data, max_val))

🎯 Contribution and Support:

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