c# - 如何将 hierarchyid 列表转换为二叉树

Question

我正在进行多层次营销(二进制)，如下所示:

(但不要求二叉树是完美的，一个节点可以有0-2个子节点)

我的问题是我从数据库中获取的数据是平面列表。

请注意我使用的是 hierarchyid (sql server 2014)

TextNode 列基本上就像一个面包屑。

每个斜杠/代表一个级别。

如果我有 /1/ 的 TextNode 作为 root。那么每个以 /1/ 开头的节点都属于那个根，即 /1/、/1/1/ 和 /1/1/1/(包含根节点，即0级)

我试过 accepted answer在这个问题上，但它不起作用。

如何将平面列表转换为二叉树，以便我可以轻松遍历并将其显示在屏幕上？

如果重要的话，我会使用 C#、ASP MVC 5、SQL Server 2014。

Answer 1

我根据 Alex 的实现准确地实现了这段代码，但正如在某些情况下提到的那样，它无法正常工作..看看我的图像和我的代码(从 Alex 的帖子中复制)[数据库中的数据是正确的但在 TreeView 中似乎有些问题]

public class Row : IRow<string>
{
    public string TextNode { get; }
    public string Value { get; }
    public long Id { get; }
    public string FIN { get; }
    public Row(string textNode, string userName, long id, string fin)
    {
        FIN = fin;
        Id = id;
        TextNode = textNode;
        Value = userName;
    }
}

public interface IRow<out T>
{
    string TextNode { get; }
    long Id { get; }
    string FIN { get; }
    T Value { get; }
}

public class TreeNode<T>
{
    private struct NodeDescriptor
    {
        public int Level { get; }
        public int ParentIndex { get; }

        public NodeDescriptor(IRow<T> row)
        {
            var split = row.TextNode.Split(new[] { "/" }, StringSplitOptions.RemoveEmptyEntries);
            Level = split.Length;
            ParentIndex = split.Length > 1 ? int.Parse(split[split.Length - 2]) - 1 : 0;
        }
    }
    public T title { get; }
    public long Id { get; }
    public string FIN { get; }
    public List<TreeNode<T>> children { get; }

    private TreeNode(T value, long id, string fin)
    {
        Id = id;
        FIN = fin;
        title = value;
        children = new List<TreeNode<T>>();
    }

    public static TreeNode<T> Parse(IReadOnlyList<IRow<T>> rows)
    {
        if (rows.Count == 0)
            return null;
        var result = new TreeNode<T>(rows[0].Value, rows[0].Id, rows[0].FIN);
        FillParents(new[] { result }, rows, 1, 1);
        return result;
    }

    private static void FillParents(IList<TreeNode<T>> parents, IReadOnlyList<IRow<T>> rows, int index, int currentLevel)
    {
        var result = new List<TreeNode<T>>();
        for (int i = index; i < rows.Count; i++)
        {
            var descriptor = new NodeDescriptor(rows[i]);
            if (descriptor.Level != currentLevel)
            {
                FillParents(result, rows, i, descriptor.Level);
                return;
            }
            var treeNode = new TreeNode<T>(rows[i].Value, rows[i].Id, rows[i].FIN);
            parents[descriptor.ParentIndex].children.Add(treeNode);
            result.Add(treeNode);
        }
    }
}

克

这也是我的 JSON 输出以获取更多信息:

{"title":"Earth","Id":32,"FIN":"FIN","children":[{"title":"Europe","Id":33,"FIN":"FIN001","children":[{"title":"France","Id":35,"FIN":"FIN001001","children":[{"title":"Paris","Id":36,"FIN":"FIN001001001","children":[]},{"title":"Brasilia","Id":41,"FIN":"FIN002001001","children":[]},{"title":"Bahia","Id":42,"FIN":"FIN002001002","children":[]}]},{"title":"Spain","Id":38,"FIN":"FIN001002","children":[{"title":"Madrid","Id":37,"FIN":"FIN001002001","children":[{"title":"Salvador","Id":43,"FIN":"FIN002001002001","children":[]}]}]},{"title":"Italy","Id":45,"FIN":"FIN001003","children":[]},{"title":"Germany","Id":48,"FIN":"FIN001004","children":[]},{"title":"test","Id":10049,"FIN":"FIN001005","children":[]}]},{"title":"South America","Id":34,"FIN":"FIN002","children":[{"title":"Brazil","Id":40,"FIN":"FIN002001","children":[{"title":"Morano","Id":47,"FIN":"FIN001003001","children":[]}]}]},{"title":"Antarctica","Id":39,"FIN":"FIN003","children":[{"title":"McMurdo Station","Id":44,"FIN":"FIN003001","children":[]}]}]}