96. 不同的二叉搜索树

解 动态规划

https://blog.youkuaiyun.com/qq_17550379/article/details/84632621
f(i)表示 当有i个节点时，二叉树结构的个数，则

• f(0) = 1
• f(1) = 1
• f(i) = f(0)*f(i-1) + f(1)*f(i-2) + …+ f(i-1)*f(0)

class Solution:
    def numTrees(self, n: int) -> int:
        
        # 动态规划
        
        f = [0 for i in range(n+1)]
        f[0] = 1
        f[1] = 1
        
        for i in range(2, n+1):
            
            for j in range(i):
                
                f[i] += f[j] * f[i-j-1]
                
        return f[n]