函数递归类型不同

胡传骏

>>> def f(x):
        if x-2 > 0 :
                result = f(x-1)+f(x-2)
        else:
                return 1

       
>>> f(20)


Traceback (most recent call last):
  File "<pyshell#37>", line 1, in <module>
    f(20)
  File "<pyshell#36>", line 3, in f
    result = f(x-1)+f(x-2)
  File "<pyshell#36>", line 3, in f
    result = f(x-1)+f(x-2)
  File "<pyshell#36>", line 3, in f
    result = f(x-1)+f(x-2)
  [Previous line repeated 14 more times]
TypeError: unsupported operand type(s) for +: 'NoneType' and 'int'

如上，问题在哪?

疾风怪盗

缺了返回值，加个return就好了

def f(x):
    if x - 2 > 0:
        result = f(x - 1) + f(x - 2)
        return result
    else:
        return 1

print(f(20))

heidern0612

if分支满足条件后进入，没有返回值。