Count the Number of Binary Search Trees present in a Binary Tree Last Updated : 29 Dec, 2022 Comments Improve Suggest changes Like Article Like Report Given a binary tree, the task is to count the number of Binary Search Trees present in it. Examples: Input: 1 / \ 2 3 / \ / \ 4 5 6 7 Output: 4Here each leaf node represents a binary search tree and there are total 4 nodes. Input: 11 / \ 8 10 / / \ 5 9 8 / \ 4 6 Output: 6 Sub-tree rooted under node 5 is a BST 5 / \ 4 6 Another BST we have is rooted under the node 8 8 / 5 / \ 4 6 Thus total 6 BSTs are present (including the leaf nodes). Approach: A Binary Tree is a Binary Search Tree if the following are true for every node x. The largest value in left subtree (of x) is smaller than value of x.The smallest value in right subtree (of x) is greater than value of x. We traverse tree in bottom-up manner. For every traversed node, we store the information of maximum and minimum of that subtree, a variable isBST to store if it is a BST and variable num_BST to store the number of Binary search tree rooted under the current node. Below is the implementation of the above approach: C++ // C++ program to count number of Binary search trees // in a given Binary Tree #include <bits/stdc++.h> using namespace std; // Binary tree node struct Node { struct Node* left; struct Node* right; int data; Node(int data) { this->data = data; this->left = NULL; this->right = NULL; } }; // Information stored in every // node during bottom up traversal struct Info { // Stores the number of BSTs present int num_BST; // Max Value in the subtree int max; // Min value in the subtree int min; // If subtree is BST bool isBST; }; // Returns information about subtree such as // number of BST's it has Info NumberOfBST(struct Node* root) { // Base case if (root == NULL) return { 0, INT_MIN, INT_MAX, true }; // If leaf node then return from function and store // information about the leaf node if (root->left == NULL && root->right == NULL) return { 1, root->data, root->data, true }; // Store information about the left subtree Info L = NumberOfBST(root->left); // Store information about the right subtree Info R = NumberOfBST(root->right); // Create a node that has to be returned Info bst; bst.min = min(root->data, (min(L.min, R.min))); bst.max = max(root->data, (max(L.max, R.max))); // If whole tree rooted under the // current root is BST if (L.isBST && R.isBST && root->data > L.max && root->data < R.min) { // Update the number of BSTs bst.isBST = true; bst.num_BST = 1 + L.num_BST + R.num_BST; } // If the whole tree is not a BST, // update the number of BSTs else { bst.isBST = false; bst.num_BST = L.num_BST + R.num_BST; } return bst; } // Driver code int main() { struct Node* root = new Node(5); root->left = new Node(9); root->right = new Node(3); root->left->left = new Node(6); root->right->right = new Node(4); root->left->left->left = new Node(8); root->left->left->right = new Node(7); cout << NumberOfBST(root).num_BST; return 0; } Java // Java program to count // number of Binary search // trees in a given Binary Tree import java.util.*; class GFG { // Binary tree node static class Node { Node left; Node right; int data; Node(int data) { this.data = data; this.left = null; this.right = null; } }; // Information stored in every // node during bottom up traversal static class Info { // Stores the number of BSTs present int num_BST; // Max Value in the subtree int max; // Min value in the subtree int min; // If subtree is BST boolean isBST; Info(int a, int b, int c, boolean d) { num_BST = a; max = b; min = c; isBST = d; } Info() { } }; // Returns information about subtree such as // number of BST's it has static Info NumberOfBST(Node root) { // Base case if (root == null) return new Info(0, Integer.MIN_VALUE, Integer.MAX_VALUE, true); // If leaf node then return // from function and store // information about the leaf node if (root.left == null && root.right == null) return new Info(1, root.data, root.data, true); // Store information about the left subtree Info L = NumberOfBST(root.left); // Store information about the right subtree Info R = NumberOfBST(root.right); // Create a node that has to be returned Info bst = new Info(); bst.min = Math.min(root.data, (Math.min(L.min, R.min))); bst.max = Math.max(root.data, (Math.max(L.max, R.max))); // If whole tree rooted under the // current root is BST if (L.isBST && R.isBST && root.data > L.max && root.data < R.min) { // Update the number of BSTs bst.isBST = true; bst.num_BST = 1 + L.num_BST + R.num_BST; } // If the whole tree is not a BST, // update the number of BSTs else { bst.isBST = false; bst.num_BST = L.num_BST + R.num_BST; } return bst; } // Driver code public static void main(String args[]) { Node root = new Node(5); root.left = new Node(9); root.right = new Node(3); root.left.left = new Node(6); root.right.right = new Node(4); root.left.left.left = new Node(8); root.left.left.right = new Node(7); System.out.print(NumberOfBST(root).num_BST); } } // This code is contributed by Arnab Kundu Python3 # Python program to count number of Binary search # trees in a given Binary Tree INT_MIN = -2**31 INT_MAX = 2**31 class newNode(): def __init__(self, data): self.data = data self.left = None self.right = None # Returns information about subtree such as # number of BST's it has def NumberOfBST(root): # Base case if (root == None): return 0, INT_MIN, INT_MAX, True # If leaf node then return from function and store # information about the leaf node if (root.left == None and root.right == None): return 1, root.data, root.data, True # Store information about the left subtree L = NumberOfBST(root.left) # Store information about the right subtree R = NumberOfBST(root.right) # Create a node that has to be returned bst = [0]*4 bst[2] = min(root.data, (min(L[2], R[2]))) bst[1] = max(root.data, (max(L[1], R[1]))) # If whole tree rooted under the # current root is BST if (L[3] and R[3] and root.data > L[1] and root.data < R[2]): # Update the number of BSTs bst[3] = True bst[0] = 1 + L[0] + R[0] # If the whole tree is not a BST, # update the number of BSTs else: bst[3] = False bst[0] = L[0] + R[0] return bst # Driver code if __name__ == '__main__': root = newNode(5) root.left = newNode(9) root.right = newNode(3) root.left.left = newNode(6) root.right.right = newNode(4) root.left.left.left = newNode(8) root.left.left.right = newNode(7) print(NumberOfBST(root)[0]) # This code is contributed by SHUBHAMSINGH10 C# using System; // C# program to count // number of Binary search // trees in a given Binary Tree public class GFG { // Binary tree node public class Node { public Node left; public Node right; public int data; public Node(int data) { this.data = data; this.left = null; this.right = null; } } // Information stored in every // node during bottom up traversal public class Info { // Stores the number of BSTs present public int num_BST; // Max Value in the subtree public int max; // Min value in the subtree public int min; // If subtree is BST public bool isBST; public Info(int a, int b, int c, bool d) { num_BST = a; max = b; min = c; isBST = d; } public Info() { } } // Returns information about subtree such as // number of BST's it has static Info NumberOfBST(Node root) { // Base case if (root == null) return new Info(0, Int32.MinValue, Int32.MaxValue, true); // If leaf node then return // from function and store // information about the leaf node if (root.left == null && root.right == null) return new Info(1, root.data, root.data, true); // Store information about the left subtree Info L = NumberOfBST(root.left); // Store information about the right subtree Info R = NumberOfBST(root.right); // Create a node that has to be returned Info bst = new Info(); bst.min = Math.Min(root.data, (Math.Min(L.min, R.min))); bst.max = Math.Max(root.data, (Math.Max(L.max, R.max))); // If whole tree rooted under the // current root is BST if (L.isBST && R.isBST && root.data > L.max && root.data < R.min) { // Update the number of BSTs bst.isBST = true; bst.num_BST = 1 + L.num_BST + R.num_BST; } // If the whole tree is not a BST, // update the number of BSTs else { bst.isBST = false; bst.num_BST = L.num_BST + R.num_BST; } return bst; } // Driver code public static void Main(string[] args) { Node root = new Node(5); root.left = new Node(9); root.right = new Node(3); root.left.left = new Node(6); root.right.right = new Node(4); root.left.left.left = new Node(8); root.left.left.right = new Node(7); Console.Write(NumberOfBST(root).num_BST); } } // This code is contributed by Shrikant13 JavaScript <script> // JavaScript program to count // number of Binary search // trees in a given Binary Tree // Binary tree node class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } // Information stored in every // node during bottom up traversal class Info { constructor(a, b, c, d) { // Stores the number of BSTs present this.num_BST = a; // Max Value in the subtree this.max = b; // Min value in the subtree this.min = c; // If subtree is BST this.isBST = d; } } // Returns information about subtree such as // number of BST's it has function NumberOfBST(root) { // Base case if (root == null) return new Info(0, -2147483648, 2147483647, true); // If leaf node then return // from function and store // information about the leaf node if (root.left == null && root.right == null) return new Info(1, root.data, root.data, true); // Store information about the left subtree var L = NumberOfBST(root.left); // Store information about the right subtree var R = NumberOfBST(root.right); // Create a node that has to be returned var bst = new Info(); bst.min = Math.min(root.data, Math.min(L.min, R.min)); bst.max = Math.max(root.data, Math.max(L.max, R.max)); // If whole tree rooted under the // current root is BST if (L.isBST && R.isBST && root.data > L.max && root.data < R.min) { // Update the number of BSTs bst.isBST = true; bst.num_BST = 1 + L.num_BST + R.num_BST; } // If the whole tree is not a BST, // update the number of BSTs else { bst.isBST = false; bst.num_BST = L.num_BST + R.num_BST; } return bst; } // Driver code var root = new Node(5); root.left = new Node(9); root.right = new Node(3); root.left.left = new Node(6); root.right.right = new Node(4); root.left.left.left = new Node(8); root.left.left.right = new Node(7); document.write(NumberOfBST(root).num_BST); // This code is contributed by rdtank. </script> Output: 4 Time Complexity: O(n), where n is the number of nodes in the given binary tree.Auxiliary Space: O(height of the tree), since the depth of the recursive tree can go up to the height of the tree. 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