计算二叉树中存在的二叉搜索树的数量

给定一个二叉树，任务是计算其中存在的二叉搜索树的数量。

例子：

Input: 

Output: 4
Here each leaf node represents a binary search tree and there are total 4 nodes.
Input:

Output: 6 
Sub-tree rooted under node 5 is a BST

Another BST we have is rooted under the node 8

Thus total 6 BSTs are present (including the leaf nodes).

方法：如果每个节点x都满足以下条件，则二叉树就是二叉搜索树。

1. (x的)左子树中的最大值小于x的值。
2. (x的)右子树中的最小值大于x的值。

我们以自下而上的方式遍历树。对于每个遍历的节点，我们存储该子树的最大值和最小值的信息；如果变量是BST，则存储变量isBST；如果变量num_BST ，则存储变量num_BST，以存储根于当前节点下的Binary搜索树的数量。
下面是上述方法的实现：

C++

1
   /  \
  2    3
 / \  / \
4   5 6  7

Java

11
     /  \
    8    10
   /    /  \
  5    9    8
 / \
4   6

Python3

5
  / \
 4   6

C#

8    
       /    
      5    
     / \
    4   6

输出：

// C++ program to count number of Binary search trees
// in a given Binary Tree
#include 
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;
}

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