我不喜欢树的方法。它使事情变得过于复杂，包括搜索或下钻，甚至ui控件填充。

我建议使用dictionary <TChild, TParent>这种非常简单的方法。这也允许节点或级别之间没有连接。

delegate void TreeVisitor<T>(T nodeData);

class NTree<T>
{
    private T data;
    private LinkedList<NTree<T>> children;

    public NTree(T data)
    {
         this.data = data;
        children = new LinkedList<NTree<T>>();
    }

    public void AddChild(T data)
    {
        children.AddFirst(new NTree<T>(data));
    }

    public NTree<T> GetChild(int i)
    {
        foreach (NTree<T> n in children)
            if (--i == 0)
                return n;
        return null;
    }

    public void Traverse(NTree<T> node, TreeVisitor<T> visitor)
    {
        visitor(node.data);
        foreach (NTree<T> kid in node.children)
            Traverse(kid, visitor);
    }
}

简单的递归实现… < 40行代码… 你只需要在类外保留一个对树根的引用， 或者在另一个类中包装它，也许重命名为TreeNode??

这是我自己的想法:

class Program
{
    static void Main(string[] args)
    {
        var tree = new Tree<string>()
            .Begin("Fastfood")
                .Begin("Pizza")
                    .Add("Margherita")
                    .Add("Marinara")
                .End()
                .Begin("Burger")
                    .Add("Cheese burger")
                    .Add("Chili burger")
                    .Add("Rice burger")
                .End()
            .End();

        tree.Nodes.ForEach(p => PrintNode(p, 0));
        Console.ReadKey();
    }

    static void PrintNode<T>(TreeNode<T> node, int level)
    {
        Console.WriteLine("{0}{1}", new string(' ', level * 3), node.Value);
        level++;
        node.Children.ForEach(p => PrintNode(p, level));
    }
}

public class Tree<T>
{
    private Stack<TreeNode<T>> m_Stack = new Stack<TreeNode<T>>();

    public List<TreeNode<T>> Nodes { get; } = new List<TreeNode<T>>();

    public Tree<T> Begin(T val)
    {
        if (m_Stack.Count == 0)
        {
            var node = new TreeNode<T>(val, null);
            Nodes.Add(node);
            m_Stack.Push(node);
        }
        else
        {
            var node = m_Stack.Peek().Add(val);
            m_Stack.Push(node);
        }

        return this;
    }

    public Tree<T> Add(T val)
    {
        m_Stack.Peek().Add(val);
        return this;
    }

    public Tree<T> End()
    {
        m_Stack.Pop();
        return this;
    }
}

public class TreeNode<T>
{
    public T Value { get; }
    public TreeNode<T> Parent { get; }
    public List<TreeNode<T>> Children { get; }

    public TreeNode(T val, TreeNode<T> parent)
    {
        Value = val;
        Parent = parent;
        Children = new List<TreeNode<T>>();
    }

    public TreeNode<T> Add(T val)
    {
        var node = new TreeNode<T>(val, this);
        Children.Add(node);
        return node;
    }
}

输出:

Fastfood
   Pizza
      Margherita
      Marinara
   Burger
      Cheese burger
      Chili burger
      Rice burger

这是我的，和艾伦·盖奇的很相似，只是在我看来更传统一点。就我而言，我使用List<T>时没有遇到任何性能问题。如果需要，切换到LinkedList是很容易的。

namespace Overby.Collections
{
    public class TreeNode<T>
    {
        private readonly T _value;
        private readonly List<TreeNode<T>> _children = new List<TreeNode<T>>();

        public TreeNode(T value)
        {
            _value = value;
        }

        public TreeNode<T> this[int i]
        {
            get { return _children[i]; }
        }

        public TreeNode<T> Parent { get; private set; }

        public T Value { get { return _value; } }

        public ReadOnlyCollection<TreeNode<T>> Children
        {
            get { return _children.AsReadOnly(); }
        }

        public TreeNode<T> AddChild(T value)
        {
            var node = new TreeNode<T>(value) {Parent = this};
            _children.Add(node);
            return node;
        }

        public TreeNode<T>[] AddChildren(params T[] values)
        {
            return values.Select(AddChild).ToArray();
        }

        public bool RemoveChild(TreeNode<T> node)
        {
            return _children.Remove(node);
        }

        public void Traverse(Action<T> action)
        {
            action(Value);
            foreach (var child in _children)
                child.Traverse(action);
        }

        public IEnumerable<T> Flatten()
        {
            return new[] {Value}.Concat(_children.SelectMany(x => x.Flatten()));
        }
    }
}

还有另一种树结构:

public class TreeNode<T> : IEnumerable<TreeNode<T>>
{

    public T Data { get; set; }
    public TreeNode<T> Parent { get; set; }
    public ICollection<TreeNode<T>> Children { get; set; }

    public TreeNode(T data)
    {
        this.Data = data;
        this.Children = new LinkedList<TreeNode<T>>();
    }

    public TreeNode<T> AddChild(T child)
    {
        TreeNode<T> childNode = new TreeNode<T>(child) { Parent = this };
        this.Children.Add(childNode);
        return childNode;
    }

    ... // for iterator details see below link
}

示例用法:

TreeNode<string> root = new TreeNode<string>("root");
{
    TreeNode<string> node0 = root.AddChild("node0");
    TreeNode<string> node1 = root.AddChild("node1");
    TreeNode<string> node2 = root.AddChild("node2");
    {
        TreeNode<string> node20 = node2.AddChild(null);
        TreeNode<string> node21 = node2.AddChild("node21");
        {
            TreeNode<string> node210 = node21.AddChild("node210");
            TreeNode<string> node211 = node21.AddChild("node211");
        }
    }
    TreeNode<string> node3 = root.AddChild("node3");
    {
        TreeNode<string> node30 = node3.AddChild("node30");
    }
}

奖金 见羽翼丰满的树与:

迭代器 搜索 Java / c#

https://github.com/gt4dev/yet-another-tree-structure

也有可能在LINQ中使用XML:

在c#中创建XML树(LINQ to XML)

在使用树时，XML是最成熟和最灵活的解决方案，而LINQ为您提供了所需的所有工具。 您的树的配置也变得更加清晰和用户友好，因为您可以简单地使用XML文件进行初始化。

如果你需要处理对象，你可以使用XML序列化:

XML序列化