您需要一个 递归/iterator 函数遍历所有分支并获取所有节点:

You need a recursive / iterator function to iterate over all the branches and get all the nodes:

public IEnumerable<Node> GetAllNodes(Node parent)               
{
    IEnumerable<Node> GetAllNodes(IEnumerable<Node> children)
    {
        foreach(var child in children)
        {
            yield return child;
            foreach(var c in GetAllNodes(child.Children))
                yield return c;
        }
    }

    yield return parent;

    foreach(var child in GetAllNodes(parent.Children))
        yield return child;
}

如果你有一棵树:

var root = new Node(5,
    new Node(3, new Node(11), new Node(12),
    new Node(2),
    new Node(4), new Node(13)),
    new Node(1, new Node(14), new Node(15),
    new Node(6, new Node(16), new Node(17)),
    new Node(7, new Node(8), new Node(9))), new Node(10));

调用函数，传递root节点，OrderByData属性:

Call the function, pass the root node, and OrderBy the Data property:

var q = GetAllNodes(root).OrderBy(x => x.Data).Select(x => x.Data);

Console.WriteLine(string.Join(", ", q));

输出为:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

***将其设为 扩展方法.

Preferably, make it an extension method for the Node type.

static class Extensions
{
    public static IEnumerable<Node> GetAllNodes(this Node parent)
    {
        IEnumerable<Node> GetAllNodes(IEnumerable<Node> children)
        {
            foreach (var child in children)
            {
                yield return child;
                foreach (var c in GetAllNodes(child.Children))
                    yield return c;
            }
        }

        yield return parent;

        foreach (var child in GetAllNodes(parent.Children))
            yield return child;
    }
}

所以你可以这样称呼它:

So you can call it as follows:

var q = root.GetAllNodes().OrderBy(x => x.Data).Select(x => x.Data);

Console.WriteLine(string.Join(", ", q));