由 H+1 个节点组成的高度为 H 的二叉搜索树的数量

给定一个正整数H ，任务是找到由前(H + 1) 个自然数作为节点值组成的高度为H的可能二叉搜索树的数量。由于计数可能非常大，请将其打印为模 10 ⁹ + 7 。

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：根据以下观察可以解决给定的问题：

• 只有(H + 1) 个节点可用于形成高度为H的二叉树。
• 除了根节点，每个节点都有两种可能性，即要么是左孩子要么是右孩子。
• 将T(H)视为高度为H的BST的数量，其中T(0) = 1且T(H) = 2 * T(H – 1) 。
• 求解上述递推关系， T(H) 的值为2 ^H 。

因此，根据上述观察，打印2 ^H的值作为由前(H + 1) 个自然数组成的高度为H的BST的总数。

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

Javascript

// Javascript program for the above approach
 
var mod = 1000000007;
 
// Function to calculate x^y
// modulo 1000000007 in O(log y)
function power(x, y)
{
    // Stores the value of x^y
    var 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
function CountBST(H)
{
 
    return power(2, H);
}
 
// Driver Code
var H = 2;
document.write( CountBST(H));
 

输出：

4

时间复杂度： O(log ₂ H)
辅助空间： O(1)

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