本文介绍了如何通过 LINQ 展平树?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
So I have simple tree:
class MyNode
{
public MyNode Parent;
public IEnumerable<MyNode> Elements;
int group = 1;
}
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?
解决方案
You can flatten a tree like this:
IEnumerable<MyNode> Flatten(IEnumerable<MyNode> e) =>
e.SelectMany(c => Flatten(c.Elements)).Concat(new[] { e });
You can then filter by group using Where(...).
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);
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);
Call this function like this:
IEnumerable<MyNode> tree = ....
var res = tree.Flatten(node => node.Elements);
If you would prefer flattening in pre-order rather than in post-order, switch around the sides of the Concat(...).
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