Python 可变对象上的递归函数

我试图在一个类中递归地构建所有路径。以下是我到目前为止的情况：

def get_paths(self, component=None, current_path=None, all_paths=None):

    # set defaults
    if component is None: 
        component = self._ground; 
        current_path = [component,]

    ##### [START] RECURSIVE PART #####

    # get parents of component
    parents = component.parents()

    # if no parents (an endpoint/leaf), add the current_path
    if not parents:
        all_paths.append(current_path)

    # if parents ==> recurse
    # note: because we're starting from the ground and getting all parents
    # we insert the parent before the current path (not after, like if we
    # were recursively getting files in a directory)
    else:
        for parent in parents:
            self.get_paths(parent, [parent,] + current_path), all_paths)

    ##### [END] RECURSIVE PART #####

    # Note that the recursion doesn't 'return' anything, it only modifies
    # the list. We still have to return the list by the method at the end.      
    return all_paths

它从“地面”开始，然后递归，直到元素没有任何父元素。我的问题是，这是否是执行递归的一种常见方法——实际上不返回“递归部分”中的任何内容，而只是修改可变元素（此处的列表），然后稍后返回结果

---

如果上述情况不理想，那么可以举什么例子来说明如何改进？或者，返回路径列表的其他方法有哪些（上述方法与

$find./

获取路径列表的方法非常相似）。

一种简单的方法是使用一个调用私有递归方法的公共“接口”方法

大致如下：

class Klass:

    _ground = 'ground'

    # Public method.
    def get_paths(self, component=None, current_path=None):
        all_paths = []
        self._get_paths(all_paths, component, current_path)  # Call private method.
        return all_paths

    # Private method.
    def _get_paths(self, all_paths, component=None, current_path=None):
        # Modifies all_paths - executed for that side-effect.    

        # set defaults
        if component is None:
            component = self._ground;
            current_path = [component,]

        # get parents of component
        parents = component.parents()

        # if no parents (an endpoint/leaf), add the current_path
        if not parents:
            all_paths.append(current_path)
        else:
            for parent in parents:
                self._get_paths(parent, [parent,] + current_path), all_paths)

---

切勿将列表（或其他可变值）用作默认参数<定义函数时，将定义一次代码>所有路径，然后对函数的后续调用将修改该列表。实际上，您只能调用该函数一次。您需要将其设置为

None

，然后检查它，并在函数体中将结果设置为

[]

。@Boris感谢您指出这一点。我了解到，当它传递

[]

而不是

无

时，尝试调试它的困难之处在于。非常感谢。另外，我用这个更改更新了这个问题，因为我正在寻找修改元素的一般结构/功能的答案。如果我们知道要传递给它什么，为什么不删除方法中的默认参数？@David542:对于私有方法可能可以这样做。