如何通过LINQ压平树? [英] How to flatten tree via LINQ?

问题描述

所以我有一个简单的树:

So I have simple tree:

class MyNode
{
 public MyNode Parent;
 public IEnumerable<MyNode> Elements;
 int group = 1;
}

我有一个 IEnumerable.我想将所有 MyNode(包括内部节点对象 (Elements))的列表作为一个平面列表 Where group ==1.如何通过 LINQ 做这样的事情?

I have a IEnumerable<MyNode>. I want to get a list of all MyNode (including inner node objects (Elements)) as one flat list Where group == 1. How to do such thing via LINQ?

推荐答案

你可以像这样展平一棵树:

You can flatten a tree like this:

IEnumerable<MyNode> Flatten(IEnumerable<MyNode> e) =>
    e.SelectMany(c => Flatten(c.Elements)).Concat(new[] { e });

然后您可以使用 Where(...) 按 group 过滤.

You can then filter by group using Where(...).

要获得一些风格点数"，请将 Flatten 转换为静态类中的扩展函数.

To earn some "points for style", convert Flatten to an extension function in a static class.

public static IEnumerable<MyNode> Flatten(this IEnumerable<MyNode> e) =>
    e.SelectMany(c => c.Elements.Flatten()).Concat(e);

为了获得更好的风格"的更多积分，将 Flatten 转换为一个通用的扩展方法，它接受一个树和一个从节点产生后代的函数:

To earn more points for "even better style", convert Flatten to a generic extension method that takes a tree and a function that produces descendants from a node:

public static IEnumerable<T> Flatten<T>(
    this IEnumerable<T> e
,   Func<T,IEnumerable<T>> f
) => e.SelectMany(c => f(c).Flatten(f)).Concat(e);

像这样调用这个函数:

IEnumerable<MyNode> tree = ....
var res = tree.Flatten(node => node.Elements);

如果您更喜欢按前序而不是后序展平，请切换 Concat(...) 的两侧.

If you would prefer flattening in pre-order rather than in post-order, switch around the sides of the Concat(...).

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