从根到选定节点的树路径-Python方式

Question

In a python application we have a tree made up of TreeNode objects, and we had to add a property on the TreeNode class that returns the path from the tree root to that node as a list. We have implemented this in a simple recursive way, however the code looks a little verbose for python (we suspect there is a terser way of expressing a simple algorithm like this in python). Does anyone know a more pythonic way of expressing this?

Here is a simplified version of our code - it is the definition of path_from_root we are looking to improve:

class TreeNode(object):

    def __init__(self, value, parent=None):
        self.value = value
        self.parent = parent

    @property
    def path_from_root(self):
        path = []
        _build_path_from_root(self, path)
        return path


def _build_path_from_root(node, path):
    if node.parent:
        _build_path_from_root(node.parent, path)
    path.append(node)

Below are some unit tests that show how path_from_root works:

class TreePathAsListTests(unittest.TestCase):

    def setUp(self):
        self.root = TreeNode(value="root")
        self.child_1 = TreeNode(value="child 1", parent=self.root)
        self.child_2 = TreeNode(value="child 2", parent=self.root)
        self.leaf_1a = TreeNode(value="leaf 1a", parent=self.child_1)

    def test_path_from_root(self):
        self.assertEquals([self.root, self.child_1, self.leaf_1a], self.leaf_1a.path_from_root)
        self.assertEquals([self.root, self.child_2], self.child_2.path_from_root)
        self.assertEquals([self.root], self.root.path_from_root)

Update: Accepted an answer that was a clear improvement, but definitely still interested in any other ways of expressing this.

Answer 1

I would do it this way:

@property
def path_from_root(self):
    if self.parent:
        return self.parent.path_from_root + [self]
    return [self]