python3 - 遍历树并获取所有叶节点兄弟集合

Question

我有一个 json 文件，它的结构就像一棵嵌套的树：

{
    "sub": [
    {
        "code": "01",
        "name": "a"
    },
    {
        "code": "02",
        "name": "b",
        "sub": [
        {
            "code": "0201",
            "name": "b1"
        },
        {
            "code": "0202",
            "name": "b2",
            "sub":[{
                "code": "020201",
                "name": "b21"
            },{
                "code": "020202",
                "name": "b22"
            }]
        }]
    },
    {
        "code": "03",
        "name": "c",
        "sub": [
        {
            "code": "0301",
            "name": "c1"
        },
        {
            "code": "0302",
            "name": "c2"
        }]
    }]
}

我想要一个算法来获取所有叶节点兄弟节点集（只需要名称属性）。在上面的例子中应该 return:

[
 ['a'],
 ['b1'],
 ['b21','b22'],
 ['c1','c2']
]

请注意，每个元素都是一个叶节点，每个组中的节点都是兄弟节点。

如何在 python3.x 中实现它？

def traverse(tree):
    #how to implement this function?

with open('demo.json') as f:
    tree = json.load(f)
    traverse(tree)

Answer 1

你可以递归地实现这个：检查当前树是否有子节点，然后收集并生成所有作为叶子的子节点的名称。然后，在每个子节点上递归。

def traverse(tree):
    if "sub" in tree:
        yield [sub["name"] for sub in tree["sub"] if "sub" not in sub]
        for sub in tree["sub"]:
            yield from traverse(sub)

python3 - 遍历树并获取所有叶节点兄弟集合

python3 - traverse tree and get all sets of leaf node siblings

algorithm

tree-traversal

python-3.x