由H + 1个节点组成的，高度为H的二叉搜索树的数量

给定正整数H ，任务是查找由第一(H + 1)个自然数组成的，高度为H的可能二进制搜索树的数量，并将其作为节点值。由于计数可能非常大，因此将其打印为10 ⁹ + 7模。

例子：

Input: H = 2
Output: 4
Explanation: All possible BSTs of height 2 consisting of 3 nodes are as follows:

Therefore, the total number of BSTs possible is 4.

Input: H = 6
Output: 64

方法：可以根据以下观察结果解决给定问题：

• 只能使用(H + 1)个节点来形成高度为H的二叉树。
• 除根节点外，每个节点都有两种可能性，即成为左子节点还是右子节点。
• 将T(H)视为高度为H的BST数，其中T(0)= 1且T(H)= 2 * T(H – 1) 。
• 解决上述递归关系， T(H)的值为2 ^H。

因此，根据上述观察结果，将2 ^H的值打印为由第一(H +1)个自然数组成的高度H的BST总数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
const int mod = 1000000007;
 
// Function to calculate x^y
// modulo 1000000007 in O(log y)
int power(long long x, unsigned int y)
{
    // Stores the value of x^y
    int res = 1;
 
    // Update x if it exceeds mod
    x = x % mod;
 
    // If x is divisible by mod
    if (x == 0)
        return 0;
 
    while (y > 0) {
 
        // If y is odd, then
        // multiply x with result
        if (y & 1)
            res = (res * x) % mod;
 
        // Divide y by 2
        y = y >> 1;
 
        // Update the value of x
        x = (x * x) % mod;
    }
 
    // Return the value of x^y
    return res;
}
 
// Function to count the number of
// of BSTs of height H consisting
// of (H + 1) nodes
int CountBST(int H)
{
 
    return power(2, H);
}
 
// Driver Code
int main()
{
    int H = 2;
    cout << CountBST(H);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
     
static int mod = 1000000007;
 
// Function to calculate x^y
// modulo 1000000007 in O(log y)
static long power(long x, int y)
{
     
    // Stores the value of x^y
    long res = 1;
 
    // Update x if it exceeds mod
    x = x % mod;
 
    // If x is divisible by mod
    if (x == 0)
        return 0;
 
    while (y > 0)
    {
         
        // If y is odd, then
        // multiply x with result
        if ((y & 1) == 1)
            res = (res * x) % mod;
 
        // Divide y by 2
        y = y >> 1;
 
        // Update the value of x
        x = (x * x) % mod;
    }
 
    // Return the value of x^y
    return res;
}
 
// Function to count the number of
// of BSTs of height H consisting
// of (H + 1) nodes
static long CountBST(int H)
{
    return power(2, H);
}
 
// Driver code
public static void main(String[] args)
{
    int H = 2;
     
    System.out.print(CountBST(H));
}
}
 
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
 
# Function to calculate x^y
# modulo 1000000007 in O(log y)
def power(x, y):
     
    mod = 1000000007
     
    # Stores the value of x^y
    res = 1
 
    # Update x if it exceeds mod
    x = x % mod
 
    # If x is divisible by mod
    if (x == 0):
        return 0
         
    while (y > 0):
         
        # If y is odd, then
        # multiply x with result
        if (y & 1):
            res = (res * x) % mod
 
        # Divide y by 2
        y = y >> 1
 
        # Update the value of x
        x = (x * x) % mod
     
    # Return the value of x^y
    return res
 
# Function to count the number of
# of BSTs of height H consisting
# of (H + 1) nodes
def CountBST(H):
     
    return power(2, H)
 
# Driver Code
H = 2
 
print(CountBST(H))
 
# This code is contributed by rohitsingh07052

C#

// C# program for the above approach
using System;
 
class GFG{
     
static int mod = 1000000007;
 
// Function to calculate x^y
// modulo 1000000007 in O(log y)
static long power(long x, int y)
{
     
    // Stores the value of x^y
    long res = 1;
 
    // Update x if it exceeds mod
    x = x % mod;
 
    // If x is divisible by mod
    if (x == 0)
        return 0;
 
    while (y > 0)
    {
         
        // If y is odd, then
        // multiply x with result
        if ((y & 1) == 1)
            res = (res * x) % mod;
 
        // Divide y by 2
        y = y >> 1;
 
        // Update the value of x
        x = (x * x) % mod;
    }
 
    // Return the value of x^y
    return res;
}
 
// Function to count the number of
// of BSTs of height H consisting
// of (H + 1) nodes
static long CountBST(int H)
{
     
    return power(2, H);
}
 
// Driver code
static void Main()
{
    int H = 2;
     
    Console.Write(CountBST(H));
}
}
 
// This code is contributed by abhinavjain194

输出：

4

时间复杂度： O(log ₂ H)
辅助空间： O(1)