# A program to check if a Binary Tree is BST or not

• Difficulty Level : Medium
• Last Updated : 11 Nov, 2022

A binary search tree (BST) is a node-based binary tree data structure that 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.
• Each node (item in the tree) has a distinct key.

BST

Recommended Practice

Naive Approach:

The idea is to for each node, check if max value in left subtree is smaller than the node and min value in right subtree greater than the node.

Follow the below steps to solve the problem:

• If the current node is null then return true
• If the value of the left child of the node is greater than or equal to the current node then return false
• If the value of the right child of the node is less than or equal to the current node then return false
• If the left subtree or the right subtree is not a BST then return false
• Else return true

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `/* A binary tree node has data, pointer to left child``and a pointer to right child */``struct` `node {``    ``int` `data;``    ``struct` `node* left;``    ``struct` `node* right;``};` `/* Helper function that allocates a new node with the``given data and NULL left and right pointers. */``struct` `node* newNode(``int` `data)``{``    ``struct` `node* node``        ``= (``struct` `node*)``malloc``(``sizeof``(``struct` `node));``    ``node->data = data;``    ``node->left = NULL;``    ``node->right = NULL;` `    ``return` `(node);``}` `int` `maxValue(``struct` `node* node)``{``    ``if` `(node == NULL) {``        ``return` `INT16_MIN;``    ``}``    ``int` `value = node->data;``    ``int` `leftMax = maxValue(node->left);``    ``int` `rightMax = maxValue(node->right);` `    ``return` `max(value, max(leftMax, rightMax));``}` `int` `minValue(``struct` `node* node)``{``    ``if` `(node == NULL) {``        ``return` `INT16_MAX;``    ``}``    ``int` `value = node->data;``    ``int` `leftMax = minValue(node->left);``    ``int` `rightMax = minValue(node->right);` `    ``return` `min(value, min(leftMax, rightMax));``}` `/* Returns true if a binary tree is a binary search tree */``int` `isBST(``struct` `node* node)``{``    ``if` `(node == NULL)``        ``return` `1;` `    ``/* false if the max of the left is > than us */``    ``if` `(node->left != NULL``        ``&& maxValue(node->left) > node->data)``        ``return` `0;` `    ``/* false if the min of the right is <= than us */``    ``if` `(node->right != NULL``        ``&& minValue(node->right) < node->data)``        ``return` `0;` `    ``/* false if, recursively, the left or right is not a BST``     ``*/``    ``if` `(!isBST(node->left) || !isBST(node->right))``        ``return` `0;` `    ``/* passing all that, it's a BST */``    ``return` `1;``}` `/* Driver code*/``int` `main()``{``    ``struct` `node* root = newNode(4);``    ``root->left = newNode(2);``    ``root->right = newNode(5);``   ``// root->right->left = newNode(7);``    ``root->left->left = newNode(1);``    ``root->left->right = newNode(3);` `    ``// Function call``    ``if` `(isBST(root))``        ``printf``(``"Is BST"``);``    ``else``        ``printf``(``"Not a BST"``);` `    ``return` `0;``}``// this code rectify by Laijyour Luv`

## C

 `#include ``#include ``#include ` `/* A binary tree node has data, pointer to left child``   ``and a pointer to right child */``struct` `node {``    ``int` `data;``    ``struct` `node* left;``    ``struct` `node* right;``};` `/* Helper function that allocates a new node with the``   ``given data and NULL left and right pointers. */``struct` `node* newNode(``int` `data)``{``    ``struct` `node* node``        ``= (``struct` `node*)``malloc``(``sizeof``(``struct` `node));``    ``node->data = data;``    ``node->left = NULL;``    ``node->right = NULL;` `    ``return` `(node);``}` `int` `maxValue(``struct` `node* node)``{``    ``if` `(node == NULL) {``        ``return` `0;``    ``}` `    ``int` `leftMax = maxValue(node->left);``    ``int` `rightMax = maxValue(node->right);` `    ``int` `value = 0;``    ``if` `(leftMax > rightMax) {``        ``value = leftMax;``    ``}``    ``else` `{``        ``value = rightMax;``    ``}` `    ``if` `(value > node->data) {``        ``value = node->data;``    ``}` `    ``return` `value;``}` `int` `minValue(``struct` `node* node)``{``    ``if` `(node == NULL) {``        ``return` `1000000000;``    ``}` `    ``int` `leftMax = minValue(node->left);``    ``int` `rightMax = minValue(node->right);` `    ``int` `value = 0;``    ``if` `(leftMax < rightMax) {``        ``value = leftMax;``    ``}``    ``else` `{``        ``value = rightMax;``    ``}` `    ``if` `(value < node->data) {``        ``value = node->data;``    ``}` `    ``return` `value;``}` `/* Returns true if a binary tree is a binary search tree */``int` `isBST(``struct` `node* node)``{``    ``if` `(node == NULL)``        ``return` `1;` `    ``/* false if the max of the left is > than us */``    ``if` `(node->left != NULL``        ``&& maxValue(node->left) > node->data)``        ``return` `0;` `    ``/* false if the min of the right is <= than us */``    ``if` `(node->right != NULL``        ``&& minValue(node->right) < node->data)``        ``return` `0;` `    ``/* false if, recursively, the left or right is not a BST``     ``*/``    ``if` `(!isBST(node->left) || !isBST(node->right))``        ``return` `0;` `    ``/* passing all that, it's a BST */``    ``return` `1;``}` `/* Driver code*/``int` `main()``{``    ``struct` `node* root = newNode(4);``    ``root->left = newNode(2);``    ``root->right = newNode(5);``    ``root->left->left = newNode(1);``    ``root->left->right = newNode(3);` `    ``// Function call``    ``if` `(isBST(root))``        ``printf``(``"Is BST"``);``    ``else``        ``printf``(``"Not a BST"``);` `    ``getchar``();``    ``return` `0;``}`

## Java

 `// Java implementation for the above approach``import` `java.io.*;` `class` `GFG {` `  ``/* A binary tree node has data, pointer to left child``        ``and a pointer to right child */``  ``static` `class` `node {``    ``int` `data;``    ``node left, right;``  ``}` `  ``/* Helper function that allocates a new node with the``        ``given data and NULL left and right pointers. */``  ``static` `node newNode(``int` `data)``  ``{``    ``node Node = ``new` `node();``    ``Node.data = data;``    ``Node.left = Node.right = ``null``;` `    ``return` `Node;``  ``}` `  ``static` `int` `maxValue(node Node)``  ``{``    ``if` `(Node == ``null``) {``      ``return` `Integer.MIN_VALUE;``    ``}``    ``int` `value = Node.data;``    ``int` `leftMax = maxValue(Node.left);``    ``int` `rightMax = maxValue(Node.right);` `    ``return` `Math.max(value, Math.max(leftMax, rightMax));``  ``}` `  ``static` `int` `minValue(node Node)``  ``{``    ``if` `(Node == ``null``) {``      ``return` `Integer.MAX_VALUE;``    ``}``    ``int` `value = Node.data;``    ``int` `leftMax = minValue(Node.left);``    ``int` `rightMax = minValue(Node.right);` `    ``return` `Math.min(value, Math.min(leftMax, rightMax));``  ``}` `  ``/* Returns true if a binary tree is a binary search tree``     ``*/``  ``static` `int` `isBST(node Node)``  ``{``    ``if` `(Node == ``null``) {``      ``return` `1``;``    ``}``    ` `    ``/* false if the max of the left is > than us */``    ``if` `(Node.left != ``null``        ``&& maxValue(Node.left) > Node.data) {``      ``return` `0``;``    ``}``    ` `    ``/* false if the min of the right is <= than us */``    ``if` `(Node.right != ``null``        ``&& minValue(Node.right) < Node.data) {``      ``return` `0``;``    ``}``    ` `    ``/* false if, recursively, the left or right is not a``         ``* BST*/``    ``if` `(isBST(Node.left) != ``1``        ``|| isBST(Node.right) != ``1``) {``      ``return` `0``;``    ``}``    ` `    ``/* passing all that, it's a BST */``    ``return` `1``;``  ``}` `  ``public` `static` `void` `main(String[] args)``  ``{``    ``node root = newNode(``4``);``    ``root.left = newNode(``2``);``    ``root.right = newNode(``5``);``    ` `    ``// root->right->left = newNode(7);``    ``root.left.left = newNode(``1``);``    ``root.left.right = newNode(``3``);` `    ``// Function call``    ``if` `(isBST(root) == ``1``) {``      ``System.out.print(``"Is BST"``);``    ``}``    ``else` `{``      ``System.out.print(``"Not a BST"``);``    ``}``  ``}``}` `// This code is contributed by lokeshmvs21.`

Output

`Is BST`

Note: It is assumed that you have helper functions minValue() and maxValue() that return the min or max int value from a non-empty tree

Time Complexity: O(N2), As we visit every node just once and our helper method also takes O(N) time, so overall time complexity becomes O(N) * O(N) = O(N2)
Auxiliary Space: O(H), Where H is the height of the binary tree, and the extra space is used due to the function call stack.

## Approach (Efficient):

The idea is to write a utility helper function isBSTUtil(struct node* node, int min, int max) that traverses down the tree keeping track of the narrowing min and max allowed values as it goes, looking at each node only once. The initial values for min and max should be INT_MIN and INT_MAX — they narrow from there.

Note: This method is not applicable if there are duplicate elements with the value INT_MIN or INT_MAX.

Follow the below steps to solve the problem:

• Call the isBstUtil function for the root node and set the minimum value as INT_MIN and the maximum value as INT_MAX
• If the current node is NULL then return true
• If the value of the node is less than the minimum value possible or greater than the maximum value possible then return false
• Call the same function for the left and the right subtree and narrow down the minimum and maximum values for these calls accordingly

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `/* A binary tree node has data,``pointer to left child and``a pointer to right child */``class` `node {``public``:``    ``int` `data;``    ``node* left;``    ``node* right;` `    ``/* Constructor that allocates``    ``a new node with the given data``    ``and NULL left and right pointers. */``    ``node(``int` `data)``    ``{``        ``this``->data = data;``        ``this``->left = NULL;``        ``this``->right = NULL;``    ``}``};` `int` `isBSTUtil(node* node, ``int` `min, ``int` `max);` `/* Returns true if the given``tree is a binary search tree``(efficient version). */``int` `isBST(node* node)``{``    ``return` `(isBSTUtil(node, INT_MIN, INT_MAX));``}` `/* Returns true if the given``tree is a BST and its values``are >= min and <= max. */``int` `isBSTUtil(node* node, ``int` `min, ``int` `max)``{``    ``/* an empty tree is BST */``    ``if` `(node == NULL)``        ``return` `1;` `    ``/* false if this node violates``    ``the min/max constraint */``    ``if` `(node->data < min || node->data > max)``        ``return` `0;` `    ``/* otherwise check the subtrees recursively,``    ``tightening the min or max constraint */``    ``return` `isBSTUtil(node->left, min, node->data - 1)``           ``&& ``// Allow only distinct values``           ``isBSTUtil(node->right, node->data + 1,``                     ``max); ``// Allow only distinct values``}` `/* Driver code*/``int` `main()``{``    ``node* root = ``new` `node(4);``    ``root->left = ``new` `node(2);``    ``root->right = ``new` `node(5);``    ``root->left->left = ``new` `node(1);``    ``root->left->right = ``new` `node(3);` `      ``// Function call``    ``if` `(isBST(root))``        ``cout << ``"Is BST"``;``    ``else``        ``cout << ``"Not a BST"``;` `    ``return` `0;``}` `// This code is contributed by rathbhupendra`

## C

 `#include ``#include ``#include ` `/* A binary tree node has data, pointer to left child``   ``and a pointer to right child */``struct` `node {``    ``int` `data;``    ``struct` `node* left;``    ``struct` `node* right;``};` `int` `isBSTUtil(``struct` `node* node, ``int` `min, ``int` `max);` `/* Returns true if the given tree is a binary search tree`` ``(efficient version). */``int` `isBST(``struct` `node* node)``{``    ``return` `(isBSTUtil(node, INT_MIN, INT_MAX));``}` `/* Returns true if the given tree is a BST and its``   ``values are >= min and <= max. */``int` `isBSTUtil(``struct` `node* node, ``int` `min, ``int` `max)``{``    ``/* an empty tree is BST */``    ``if` `(node == NULL)``        ``return` `1;` `    ``/* false if this node violates the min/max constraint */``    ``if` `(node->data < min || node->data > max)``        ``return` `0;` `    ``/* otherwise check the subtrees recursively,``     ``tightening the min or max constraint */``    ``return` `isBSTUtil(node->left, min, node->data - 1)``           ``&& ``// Allow only distinct values``           ``isBSTUtil(node->right, node->data + 1,``                     ``max); ``// Allow only distinct values``}` `/* Helper function that allocates a new node with the``   ``given data and NULL left and right pointers. */``struct` `node* newNode(``int` `data)``{``    ``struct` `node* node``        ``= (``struct` `node*)``malloc``(``sizeof``(``struct` `node));``    ``node->data = data;``    ``node->left = NULL;``    ``node->right = NULL;` `    ``return` `(node);``}` `/* Driver code*/``int` `main()``{``    ``struct` `node* root = newNode(4);``    ``root->left = newNode(2);``    ``root->right = newNode(5);``    ``root->left->left = newNode(1);``    ``root->left->right = newNode(3);` `      ``// Function call``    ``if` `(isBST(root))``        ``printf``(``"Is BST"``);``    ``else``        ``printf``(``"Not a BST"``);` `    ``getchar``();``    ``return` `0;``}`

## Java

 `// Java implementation to check if given Binary tree``// is a BST or not` `/* Class containing left and right child of current`` ``node and key value*/``class` `Node {``    ``int` `data;``    ``Node left, right;` `    ``public` `Node(``int` `item)``    ``{``        ``data = item;``        ``left = right = ``null``;``    ``}``}` `public` `class` `BinaryTree {``    ``// Root of the Binary Tree``    ``Node root;` `    ``/* Can give min and max value according to your code or``    ``can write a function to find min and max value of tree.``  ``*/` `    ``/* Returns true if given search tree is binary``     ``search tree (efficient version) */``    ``boolean` `isBST()``    ``{``        ``return` `isBSTUtil(root, Integer.MIN_VALUE,``                         ``Integer.MAX_VALUE);``    ``}` `    ``/* Returns true if the given tree is a BST and its``      ``values are >= min and <= max. */``    ``boolean` `isBSTUtil(Node node, ``int` `min, ``int` `max)``    ``{``        ``/* an empty tree is BST */``        ``if` `(node == ``null``)``            ``return` `true``;` `        ``/* false if this node violates the min/max``         ``* constraints */``        ``if` `(node.data < min || node.data > max)``            ``return` `false``;` `        ``/* otherwise check the subtrees recursively``        ``tightening the min/max constraints */``        ``// Allow only distinct values``        ``return` `(``            ``isBSTUtil(node.left, min, node.data - ``1``)``            ``&& isBSTUtil(node.right, node.data + ``1``, max));``    ``}` `    ``/* Driver code */``    ``public` `static` `void` `main(String args[])``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(``4``);``        ``tree.root.left = ``new` `Node(``2``);``        ``tree.root.right = ``new` `Node(``5``);``        ``tree.root.left.left = ``new` `Node(``1``);``        ``tree.root.left.right = ``new` `Node(``3``);` `          ``// Function call``        ``if` `(tree.isBST())``            ``System.out.println(``"Is BST"``);``        ``else``            ``System.out.println(``"Not a BST"``);``    ``}``}`

## Python3

 `# Python program to check if a binary tree is bst or not` `INT_MAX ``=` `4294967296``INT_MIN ``=` `-``4294967296` `# A binary tree node`  `class` `Node:` `    ``# Constructor to create a new node``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None`  `# Returns true if the given tree is a binary search tree``# (efficient version)``def` `isBST(node):``    ``return` `(isBSTUtil(node, INT_MIN, INT_MAX))` `# Returns true if the given tree is a BST and its values``# >= min and <= max`  `def` `isBSTUtil(node, mini, maxi):` `    ``# An empty tree is BST``    ``if` `node ``is` `None``:``        ``return` `True` `    ``# False if this node violates min/max constraint``    ``if` `node.data < mini ``or` `node.data > maxi:``        ``return` `False` `    ``# Otherwise check the subtrees recursively``    ``# tightening the min or max constraint``    ``return` `(isBSTUtil(node.left, mini, node.data ``-` `1``) ``and``            ``isBSTUtil(node.right, node.data``+``1``, maxi))`  `# Driver code``if` `__name__ ``=``=` `"__main__"``:``  ``root ``=` `Node(``4``)``  ``root.left ``=` `Node(``2``)``  ``root.right ``=` `Node(``5``)``  ``root.left.left ``=` `Node(``1``)``  ``root.left.right ``=` `Node(``3``)` `  ``# Function call``  ``if` `(isBST(root)):``      ``print``(``"Is BST"``)``  ``else``:``      ``print``(``"Not a BST"``)` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007)`

## C#

 `using` `System;` `// C# implementation to check if given Binary tree``// is a BST or not` `/* Class containing left and right child of current`` ``node and key value*/``public` `class` `Node {``    ``public` `int` `data;``    ``public` `Node left, right;` `    ``public` `Node(``int` `item)``    ``{``        ``data = item;``        ``left = right = ``null``;``    ``}``}` `public` `class` `BinaryTree {``    ``// Root of the Binary Tree``    ``public` `Node root;` `    ``/* can give min and max value according to your code or``    ``can write a function to find min and max value of tree.``  ``*/` `    ``/* returns true if given search tree is binary``     ``search tree (efficient version) */``    ``public` `virtual` `bool` `BST``    ``{``        ``get``        ``{``            ``return` `isBSTUtil(root, ``int``.MinValue,``                             ``int``.MaxValue);``        ``}``    ``}` `    ``/* Returns true if the given tree is a BST and its``      ``values are >= min and <= max. */``    ``public` `virtual` `bool` `isBSTUtil(Node node, ``int` `min,``                                  ``int` `max)``    ``{``        ``/* an empty tree is BST */``        ``if` `(node == ``null``) {``            ``return` `true``;``        ``}` `        ``/* false if this node violates the min/max``         ``* constraints */``        ``if` `(node.data < min || node.data > max) {``            ``return` `false``;``        ``}` `        ``/* otherwise check the subtrees recursively``        ``tightening the min/max constraints */``        ``// Allow only distinct values``        ``return` `(``            ``isBSTUtil(node.left, min, node.data - 1)``            ``&& isBSTUtil(node.right, node.data + 1, max));``    ``}` `    ``/* Driver code */``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``BinaryTree tree = ``new` `BinaryTree();``        ``tree.root = ``new` `Node(4);``        ``tree.root.left = ``new` `Node(2);``        ``tree.root.right = ``new` `Node(5);``        ``tree.root.left.left = ``new` `Node(1);``        ``tree.root.left.right = ``new` `Node(3);``      ` `        ``// Function call``        ``if` `(tree.BST) {``            ``Console.WriteLine(``"IS BST"``);``        ``}``        ``else` `{``            ``Console.WriteLine(``"Not a BST"``);``        ``}``    ``}``}` `// This code is contributed by Shrikant13`

## Javascript

 `// Javascript implementation to``// check if given Binary tree``// is a BST or not`` ` `/* Class containing left and right child of current`` ``node and key value*/` `class Node``{``    ``constructor(item)``    ``{``        ``this``.data=item;``        ``this``.left=``this``.right=``null``;``    ``}``}` ` ``//Root of the Binary Tree``    ``let root;``    ` `    ``/* can give min and max value according to your code or``    ``can write a function to find min and max value of tree. */`` ` `    ``/* returns true if given search tree is binary``     ``search tree (efficient version) */``    ``function` `isBST()``    ``{``        ``return` `isBSTUtil(root, Number.MIN_VALUE,``                               ``Number.MAX_VALUE);``    ``}``    ` `    ``/* Returns true if the given tree is a BST and its``      ``values are >= min and <= max. */``    ``function` `isBSTUtil(node,min,max)``    ``{``        ``/* an empty tree is BST */``        ``if` `(node == ``null``)``            ``return` `true``;`` ` `        ``/* false if this node violates``        ``the min/max constraints */``        ``if` `(node.data < min || node.data > max)``            ``return` `false``;`` ` `        ``/* otherwise check the subtrees recursively``        ``tightening the min/max constraints */``        ``// Allow only distinct values``        ``return` `(isBSTUtil(node.left, min, node.data-1) &&``                ``isBSTUtil(node.right, node.data+1, max));``    ``}``    ` `     ``/* Driver program to test above functions */``        ``root = ``new` `Node(4);``        ``root.left = ``new` `Node(2);``        ``root.right = ``new` `Node(5);``        ``root.left.left = ``new` `Node(1);``        ``root.left.right = ``new` `Node(3);`` ` `        ``if` `(isBST())``            ``document.write(``"IS BST
"``);``        ``else``            ``document.write(``"Not a BST
"``);` `// This code is contributed by rag2127`

Output

`Is BST`

Time Complexity: O(N), Where N is the number of nodes in the tree
Auxiliary Space: O(1), if Function Call Stack size is not considered, otherwise O(H) where H is the height of the tree

## Check whether the binary tree is BST or not using inorder traversal:

The idea is to use Inorder traversal of a binary search tree generates output, sorted in ascending order. So generate inorder traversal of the  given binary tree and check if the values are sorted or not

Follow the below steps to solve the problem:

• Do In-Order Traversal of the given tree and store the result in a temp array.
• This method assumes that there are no duplicate values in the tree
• Check if the temp array is sorted in ascending order, if it is, then the tree is BST.

Note: We can avoid the use of an Auxiliary Array. While doing In-Order traversal, we can keep track of previously visited nodes. 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 to``left child and a pointer to right child */``struct` `Node {``    ``int` `data;``    ``struct` `Node *left, *right;` `    ``Node(``int` `data)``    ``{``        ``this``->data = data;``        ``left = right = NULL;``    ``}``};` `bool` `isBSTUtil(``struct` `Node* root, Node*& 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` `(prev != NULL && root->data <= prev->data)``            ``return` `false``;` `        ``prev = root;` `        ``return` `isBSTUtil(root->right, prev);``    ``}` `    ``return` `true``;``}` `bool` `isBST(Node* root)``{``    ``Node* prev = NULL;``    ``return` `isBSTUtil(root, prev);``}` `/* Driver code*/``int` `main()``{``    ``struct` `Node* root = ``new` `Node(3);``    ``root->left = ``new` `Node(2);``    ``root->right = ``new` `Node(5);``    ``root->left->left = ``new` `Node(1);``    ``root->left->right = ``new` `Node(4);` `    ``// Function call``    ``if` `(isBST(root))``        ``cout << ``"Is BST"``;``    ``else``        ``cout << ``"Not a BST"``;` `    ``return` `0;``}`

## Java

 `// Java program to check if a given tree is BST.``import` `java.io.*;` `class` `GFG {``    ``/* A binary tree node has data, pointer to``    ``left child and a pointer to right child */``    ``public` `static` `class` `Node {``        ``public` `int` `data;``        ``public` `Node left, right;` `        ``public` `Node(``int` `data)``        ``{``            ``this``.data = data;``            ``left = right = ``null``;``        ``}``    ``};` `    ``static` `Node prev;` `    ``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` `(prev != ``null` `&& root.data <= prev.data)``                ``return` `false``;` `            ``prev = root;` `            ``return` `isBSTUtil(root.right);``        ``}``        ``return` `true``;``    ``}` `    ``static` `Boolean isBST(Node root)``    ``{``        ``return` `isBSTUtil(root);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``Node root = ``new` `Node(``3``);``        ``root.left = ``new` `Node(``2``);``        ``root.right = ``new` `Node(``5``);``        ``root.left.left = ``new` `Node(``1``);``        ``root.left.right = ``new` `Node(``4``);` `        ``// Function call``        ``if` `(isBST(root))``            ``System.out.println(``"Is BST"``);``        ``else``            ``System.out.println(``"Not a BST"``);``    ``}``}` `// This code is contributed by Shubham Singh`

## Python3

 `# Python3 program to check``# if a given tree is BST.``import` `math` `# A binary tree node has data,``# pointer to left child and``# a pointer to right child``class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None`  `def` `isBSTUtil(root, prev):` `    ``# traverse the tree in inorder fashion``    ``# and keep track of prev node``    ``if` `(root !``=` `None``):``        ``if` `(isBSTUtil(root.left, prev) ``=``=` `True``):``            ``return` `False` `        ``# Allows only distinct valued nodes``        ``if` `(prev !``=` `None` `and``                ``root.data <``=` `prev.data):``            ``return` `False` `        ``prev ``=` `root``        ``return` `isBSTUtil(root.right, prev)` `    ``return` `True`  `def` `isBST(root):``    ``prev ``=` `None``    ``return` `isBSTUtil(root, prev)`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``root ``=` `Node(``3``)``    ``root.left ``=` `Node(``2``)``    ``root.right ``=` `Node(``5``)``    ``root.right.left ``=` `Node(``1``)``    ``root.right.right ``=` `Node(``4``)` `    ``# Function call``    ``if` `(isBST(root) ``=``=` `None``):``        ``print``(``"Is BST"``)``    ``else``:``        ``print``(``"Not a BST"``)` `# This code is contributed by Srathore`

## C#

 `// C# program to check if a given tree is BST.``using` `System;``public` `class` `GFG {``    ``/* A binary tree node has data, pointer to``    ``left child and a pointer to right child */``    ``public` `class` `Node {``        ``public` `int` `data;``        ``public` `Node left, right;` `        ``public` `Node(``int` `data)``        ``{``            ``this``.data = data;``            ``left = right = ``null``;``        ``}``    ``};` `    ``static` `Node prev;` `    ``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` `(prev != ``null` `&& root.data <= prev.data)``                ``return` `false``;` `            ``prev = root;` `            ``return` `isBSTUtil(root.right);``        ``}``        ``return` `true``;``    ``}` `    ``static` `Boolean isBST(Node root)``    ``{``        ``return` `isBSTUtil(root);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``Node root = ``new` `Node(3);``        ``root.left = ``new` `Node(2);``        ``root.right = ``new` `Node(5);``        ``root.left.left = ``new` `Node(1);``        ``root.left.right = ``new` `Node(4);` `        ``// Function call``        ``if` `(isBST(root))``            ``Console.WriteLine(``"Is BST"``);``        ``else``            ``Console.WriteLine(``"Not a BST"``);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 `// Javascript program to check if a given tree is BST.   ``    ``class Node``    ``{``        ``constructor(data)``        ``{``           ``this``.left = ``null``;``           ``this``.right = ``null``;``           ``this``.data = data;``        ``}``    ``}``    ` `    ``let prev;`` ` `    ``function` `isBSTUtil(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` `(prev != ``null` `&& root.data <= prev.data)``                ``return` `false``;` `            ``prev = root;` `            ``return` `isBSTUtil(root.right);``        ``}``        ``return` `true``;``    ``}` `    ``function` `isBST(root)``    ``{``        ``return` `isBSTUtil(root);``    ``}``    ` `    ``let root = ``new` `Node(3);``    ``root.left = ``new` `Node(2);``    ``root.right = ``new` `Node(5);``    ``root.left.left = ``new` `Node(1);``    ``root.left.right = ``new` `Node(4);`` ` `    ``if` `(isBST(root))``        ``document.write(``"Is BST"``);``    ``else``        ``document.write(``"Not a BST"``);` `// This code is contributed by divyeshrabadiya07.`

Output

`Not a BST`

Time Complexity: O(N), Where N is the number of nodes in the tree
Auxiliary Space: O(H), Here H is the height of the tree and the extra space is used due to the function call stack.

Please write comments if you find any bug in the above programs/algorithms or other ways to solve the same problem.

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