用递归法求解，给定节点个数求二叉树的种类（python)

(1)0个节点，也是一种二叉树
(2)二叉树的种类 = 左子树种类数 * 右子树种类数
# 递归表达式
(3)bt(3) = bt(0) * bt(2) + bt(1) * bt(1) + bt(2) * bt(0)
(4)bt(n) = bt(0) * bt(n-1) + bt(1) * bt(n-2) + bt(2) * bt (n-3) +.....+ bt(0) * bt(n-1)
# 递归边界
(5)bt(0) = 1    bt(1) = 1

def binary_tree(n, buffer):
   if n in (0, 1):
       buffer[n] = 1
       return 1
   elif n in buffer:
       return buffer[n]
   else:
       result = 0
       for i in range(n):
           result += binary_tree(i, buffer) * binary_tree(n-i-1, buffer)
       buffer[n] =result
       return result

if __name__ == '__main__':
    buffer = {}
    for i in range(10):
        print(binary_tree(i, buffer))
    print(buffer)