# Check if a Binary Tree is BST : Simple and Efficient Approach

Given a Binary Tree, the task is to check whether the given binary tree is Binary Search Tree or not.
A binary search tree (BST) is a node-based binary tree data structure which has the following properties.

• 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.

From the above properties it naturally follows that:

• Each node (item in the tree) has a distinct key.

We have already discussed different approaches to solve this problem in the previous article.
In this article, we will discuss a simple yet efficient approach to solve the above problem.
The idea is to use Inorder traversal and keep track of the previously visited node’s value. Since the inorder traversal of a BST generates a sorted array as output, So, the previous element should always be less than or equals to the current element.
While doing In-Order traversal, we can keep track of previously visited Node’s value and if the value of the currently visited node is less than the previous value, then the tree is not BST.

Below is the implementation of the above approach:

## C++

 // C++ program to check if a given tree is BST.#include using namespace std; /* A binary tree node has data, pointer toleft child and a pointer to right child */struct Node {    int data;    struct Node *left, *right;     Node(int data)    {        this->data = data;        left = right = NULL;    }}; // Utility function to check if Binary Tree is BSTbool isBSTUtil(struct Node* root, int& prev){    // traverse the tree in inorder fashion and    // keep track of prev node    if (root) {        if (!isBSTUtil(root->left, prev))            return false;         // Allows only distinct valued nodes        if (root->data <= prev)            return false;         // Initialize prev to current        prev = root->data;         return isBSTUtil(root->right, prev);    }     return true;} // Function to check if Binary Tree is BSTbool isBST(Node* root){    int prev = INT_MIN;    return isBSTUtil(root, prev);} /* Driver code*/int main(){    struct Node* root = new Node(5);    root->left = new Node(2);    root->right = new Node(15);    root->left->left = new Node(1);    root->left->right = new Node(4);     if (isBST(root))        cout << "Is BST";    else        cout << "Not a BST";     return 0;}

## Java

 // Java program to check if a given tree is BST.class GFG {     static int prev = Integer.MIN_VALUE;    /* A binary tree node has data, pointer to    left child and a pointer to right child */    static class Node {        int data;        Node left, right;         Node(int data)        {            this.data = data;            left = right = null;        }    };     // Utility function to check if Binary Tree is BST    static boolean isBSTUtil(Node root)    {        // traverse the tree in inorder fashion and        // keep track of prev node        if (root != null) {            if (!isBSTUtil(root.left))                return false;             // Allows only distinct valued nodes            if (root.data <= prev)                return false;             // Initialize prev to current            prev = root.data;             return isBSTUtil(root.right);        }         return true;    }     // Function to check if Binary Tree is BST    static boolean isBST(Node root)    {        return isBSTUtil(root);    }     /* Driver code*/    public static void main(String[] args)    {        Node root = new Node(5);        root.left = new Node(2);        root.right = new Node(15);        root.left.left = new Node(1);        root.left.right = new Node(4);         if (isBST(root))            System.out.print("Is BST");        else            System.out.print("Not a BST");    }} // This code is contributed by PrinciRaj1992

## Python3

 # Python3 program to check if a given tree is BST. import mathprev = -math.inf  class Node:    """    Creates a Binary tree node that has data,     a pointer to it's left and right child    """     def __init__(self, data):        self.left = None        self.right = None        self.data = data  def checkBST(root):    """    Function to check if Binary Tree is    a Binary Search Tree    :param root: current root node    :return: Boolean value    """    # traverse the tree in inorder     # fashion and update the prev node    global prev     if root:        if not checkBST(root.left):            return False         # Handles same valued nodes        if root.data < prev:            return False         # Set value of prev to current node        prev = root.data         return checkBST(root.right)    return True # Driver Codedef main():    root = Node(1)    root.left = Node(2)    root.right = Node(15)    root.left.left = Node(1)    root.left.right = Node(4)     if checkBST(root):        print("Is BST")    else:        print("Not a BST")  if __name__ == '__main__':    main() # This code is contributed by priyankapunjabi94

## C#

 // C# program to check if a given tree is BST.using System; class GFG {     /* A binary tree node has data, pointer to    left child and a pointer to right child */    class Node {        public int data;        public Node left, right;         public Node(int data)        {            this.data = data;            left = right = null;        }    };     // Utility function to check if Binary Tree is BST    static bool isBSTUtil(Node root, int prev)    {        // traverse the tree in inorder fashion and        // keep track of prev node        if (root != null) {            if (!isBSTUtil(root.left, prev))                return false;             // Allows only distinct valued nodes            if (root.data <= prev)                return false;             // Initialize prev to current            prev = root.data;             return isBSTUtil(root.right, prev);        }         return true;    }     // Function to check if Binary Tree is BST    static bool isBST(Node root)    {        int prev = int.MinValue;        return isBSTUtil(root, prev);    }     /* Driver code*/    public static void Main(String[] args)    {        Node root = new Node(5);        root.left = new Node(2);        root.right = new Node(15);        root.left.left = new Node(1);        root.left.right = new Node(4);         if (isBST(root))            Console.Write("Is BST");        else            Console.Write("Not a BST");    }} // This code is contributed by PrinciRaj1992

## Javascript



Output
Is BST

Time Complexity: O(N)
Auxiliary Space: O(1)

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