如何返回递归树中的所有叶子

Question

我正在尝试创建一个函数，该函数将返回递归树中的所有叶子。我看到了很多关于它的其他帖子，但我无法将其修改为我自己的代码。我正在尝试像决策树一样。这是我的代码：

class Node:
def __init__(self, data, positive_child=None, negative_child=None):
    self.data = data
    self.positive_child = positive_child
    self.negative_child = negative_child
    self.children_list = []

class Decision:
    def __init__(self, root: Node):
        self.root = root
        self.current = root

    def collect_leaves(self, node, leafs):
        if node is not None:
            if len(node.children_list) == 0:
                leafs.append(node.data)
            for n in node.children_list:
                self.collect_leaves(n, leafs)

    def return_all_leaves(self):
        leafs = []
        self.collect_leaves(self.root, leafs)
        return leafs

由于某种原因，它只返回根，而不是叶子..

例如：

flu_leaf2 = Node("influenza", None, None)
cold_leaf2 = Node("cold", None, None)
hard_leaf2 = Node("hard influenza", None, None)
headache_node2 = Node("headache", hard_leaf2, flu_leaf2)
inner_vertex2 = Node("fever", headache_node2, cold_leaf2)
healthy_leaf2 = Node("healthy", None, None)
root2 = Node("cough", inner_vertex2, healthy_leaf2)
diagnoser2 = Diagnoser(root2)

diagnoser2.return_all_leaves(self) 应该返回：

['hard influenza', 'influenza','cold','healthy']

Answer 1

我不确定您是否需要children_list。这似乎只是一个额外的维护。我认为确定节点是否具有positive_child 或negative_child 应该就足够了。

注意向 Node 添加了 __str__ 方法，以便在打印时显示一些不错的东西...

看看：

class Node:
    def __init__(self, data, positive_child=None, negative_child=None):
        self.data = data
        self.positive_child = positive_child
        self.negative_child = negative_child

    def __str__(self) -> str:
        return self.data

class Decision:
    def __init__(self, root: Node):
        self.root = root

    def collect_leaves(self, node):
        ## ---------------------------
        ## This node has no children. It is a leaf
        ## ---------------------------
        if not node.positive_child and not node.negative_child:
            return [node]
        ## ---------------------------

        ## ---------------------------
        ## Recursively collect the leaves of children
        ## ---------------------------
        leaves = []
        if node.positive_child:
            leaves.extend(self.collect_leaves(node.positive_child))
        if node.negative_child:
            leaves.extend(self.collect_leaves(node.negative_child))
        return leaves
        ## ---------------------------

    def return_all_leaves(self):
        return self.collect_leaves(self.root)

my_decsion = Decision(
    Node(
        "root",
        Node("root_a", None, None),
        Node(
            "root_b",
            Node("root_b_1", None),
            Node("root_b_2", None),
        ),
    )
)

for node in my_decsion.return_all_leaves():
    print(node)

这应该给你：

root_a
root_b_1
root_b_2

请注意，虽然 python 中的递归有点受限，但您可能希望查看不基于它的实现。

如果您想要一个不基于递归的return_all_leaves() 版本，您可以尝试：

    def return_all_leaves2(self):
        leaves = []
        todo = [self.root]
        while todo:
            this_node = todo.pop(0)

            ## ---------------------------
            ## This node was None... I think this is cleaner than testing
            ## parent_node.positive_child and parent_node.negative_child
            ## ---------------------------
            if not this_node:
                continue
            ## ---------------------------

            ## ---------------------------
            ## This node has no children. It is a leaf
            ## ---------------------------
            if not this_node.positive_child and not this_node.negative_child:
                leaves.append(this_node)
                continue
            ## ---------------------------

            ## ---------------------------
            ## add the leaves of children to future work
            ## ---------------------------
            todo.append(this_node.positive_child)
            todo.append(this_node.negative_child)
            ## ---------------------------

        return leaves

如果我们使用您的数据进行测试：

my_decsion2 = Decision(
    Node(
        "cough",
        Node(
            "fever",
            Node(
                "headache",
                Node("hard influenza", None, None),
                Node("influenza", None, None)
            ),
            Node("cold", None, None)
        ),
        Node("healthy", None, None)
    )
)

for node in my_decsion2.return_all_leaves():
    print(node)

此代码打印：

hard influenza
influenza
cold
healthy