计算高度为 h 的平衡二叉树

给定高度 h，计算并返回高度为 h 的平衡二叉树的最大数量。平衡二叉树是其中每个节点的左子树和右子树的高度差不超过1的树。
例子 ：

Input : h = 3
Output : 15

Input : h = 4
Output : 315

以下是高度为 3 的平衡二叉树。

树的高度，h = 1 + max(左高，右高)

由于左右子树的高度差不超过1，因此左右部分可能的高度可以是以下之一：

1. (h-1), (h-2)
2. (h-2), (h-1)
3. (h-1), (h-1)

count(h) = count(h-1) * count(h-2) + 
           count(h-2) * count(h-1) + 
           count(h-1) * count(h-1)
        = 2 * count(h-1) * count(h-2) +  
          count(h-1) * count(h-1)
       = count(h-1) * (2*count(h - 2) + 
                          count(h - 1))

因此我们可以看出该问题具有最优子结构性质。
计算高度为 h 的平衡二叉树的数量的递归函数是：

int countBT(int h)
{
    // One tree is possible with height 0 or 1
    if (h == 0 || h == 1)
        return 1;
    return countBT(h-1) * (2 *countBT(h-2) +
                              countBT(h-1));
}

这种递归方法的时间复杂度将是指数级的。 h = 3 问题的递归树如下所示：

正如我们所见，子问题被反复解决。因此，我们在计算结果时存储结果。
一种有效的动态规划方法如下：
下面是上述方法的实现：

C++

// C++ program to count number of balanced
// binary trees of height h.
#include 
#define mod 1000000007
using namespace std;
  
long long int countBT(int h) {
      
    long long int dp[h + 1];
    //base cases
    dp[0] = dp[1] = 1;
    for(int i = 2; i <= h; i++) {
        dp[i] = (dp[i - 1] * ((2 * dp [i - 2])%mod + dp[i - 1])%mod) % mod;
    }
    return dp[h];
}
 
 
// Driver program
int main()
{
    int h = 3;
    cout << "No. of balanced binary trees"
            " of height h is: "
         << countBT(h) << endl;
}

Java

// Java program to count number of balanced
// binary trees of height h.
class GFG {
     
    static final int MOD = 1000000007;
     
    public static long countBT(int h) {
        long[] dp = new long[h + 1];
         
        // base cases
        dp[0] = 1;
        dp[1] = 1;
         
        for(int i = 2; i <= h; ++i)
            dp[i] = (dp[i - 1] * ((2 * dp [i - 2])% MOD + dp[i - 1]) % MOD) % MOD;
             
            return dp[h];
    }
     
    // Driver program
    public static void main (String[] args) {
        int h = 3;
        System.out.println("No. of balanced binary trees of height "+h+" is: "+countBT(h));
    }
}
/*
This code is contributed by
Brij Raj Kishore
*/

Python3

# Python3 program to count number of balanced
# binary trees of height h.
 
def countBT(h) :
    MOD = 1000000007
    #initialize list
    dp = [0 for i in range(h + 1)]
     
    #base cases
    dp[0] = 1
    dp[1] = 1
     
    for i in range(2, h + 1) :
        dp[i] = (dp[i - 1] * ((2 * dp [i - 2])%MOD + dp[i - 1])%MOD) % MOD
     
    return dp[h]
 
#Driver program
h = 3
print("No. of balanced binary trees of height "+str(h)+" is: "+str(countBT(h)))
 
# This code is contributed by
# Brij Raj Kishore

C#

// C# program to count number of balanced
// binary trees of height h.
 
using System;
class GFG {
     
    static int MOD = 1000000007;
     
    public static long countBT(int h) {
        long[] dp = new long[h + 1];
         
        // base cases
        dp[0] = 1;
        dp[1] = 1;
         
        for(int i = 2; i <= h; ++i)
            dp[i] = (dp[i - 1] * ((2 * dp [i - 2])% MOD + dp[i - 1]) % MOD) % MOD;
             
            return dp[h];
    }
     
    // Driver program
    static void Main () {
        int h = 3;
        Console.WriteLine("No. of balanced binary trees of height "+h+" is: "+countBT(h));
    }
    // This code is contributed by Ryuga
}

Javascript

    // Javascript program to count number of balanced binary trees of height h.
     
    let MOD = 1000000007;
       
    function countBT(h) {
        let dp = new Array(h + 1);
        dp.fill(0);
           
        // base cases
        dp[0] = 1;
        dp[1] = 1;
           
        for(let i = 2; i <= h; ++i)
            dp[i] = (dp[i - 1] * ((2 * dp [i - 2])% MOD + dp[i - 1]) % MOD) % MOD;
               
            return dp[h];
    }
     
    let h = 3;
      document.write("No. of balanced binary trees of height h is: "+countBT(h));
     

输出 ：

No of balanced binary trees of height h is: 15

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