Python 树的分支列表 [英] List of branches of a Python tree

问题描述

我有一棵 Python 对象树.树是内在定义的:每个对象都有一个子项列表(可能为空).我希望能够打印从根到每个叶子的所有路径的列表.

I have a tree of Python objects. The tree is defined intrinsically: each object has a list (potentially empty) of children. I would like to be able to print a list of all paths from the root to each leaf.

在上面的树的情况下，这意味着:

In the case of the tree above, this would mean:

result = [
          [Node_0001,Node_0002,Node_0004],
          [Node_0001,Node_0002,Node_0005,Node_0007],
          [Node_0001,Node_0003,Node_0006],
         ]

必须将节点视为对象而不是整数(仅显示其整数 ID).我不关心结果中分支的顺序.每个节点有任意数量的子节点，递归级别也不固定.

The nodes must be treated as objects and not as integers (only their integer ID is displayed). I don't care about the order of branches in the result. Each node has an arbitrary number of children, and the level of recursion is not fixed either.

我正在尝试递归方法:

def get_all_paths(node):
    if len(node.children)==0:
        return [[node]]
    else:
        return [[node] + get_all_paths(child) for child in node.children]

但我最终得到了嵌套列表，这不是我想要的:

but I end-up with nested lists, which is not what I want:

[[Node_0001,
  [Node_0002, [Node_0004]],
  [Node_0002, [Node_0005, [Node_0007]]]],
 [Node_0001, [Node_0003, [Node_0006]]]]

欢迎任何帮助，这个问题让我发疯:p

谢谢

推荐答案

我认为这就是您正在尝试的:

I think this is what you are trying:

def get_all_paths(node):
    if len(node.children) == 0:
        return [[node]]
    return [
        [node] + path for child in node.children for path in get_all_paths(child)
    ]

对于节点的每个子节点，您应该获取子节点的所有路径并将节点本身添加到每个路径.您将节点添加到路径列表中，而不是单独添加每个路径.

For each child of a node, you should take all paths of the child and prepend the node itself to each path. You prepended the node to the list of paths, not every path individually.

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