如何用C#/.NET将N叉目录树转换为带路径的文件夹文件列表

我有一个N叉树结构，节点分为**文件夹（Folder）和文件（File，叶子节点）**两种类型。文件夹节点可以包含子文件夹或文件节点。需要将该结构转换为一个列表，每个条目包含从根到目标文件夹的复合路径，以及该文件夹下的所有文件节点。

示例结构 #

• Root
  • Folder1
    • File1
    • File2
  • Folder2
    • Folder3
      • File3
    • File4

期望转换结果 #

1. Folder1 - File1, File2
2. Folder2.Folder3 - File3
3. Folder2 - File4

初始节点结构代码 #

public interface INode
{
}

public class Folder : INode
{
    public string Name { get; set; }
    public List<INode>? Children { get; set; }
}

public class File : INode
{
    public string Name { get; set; }
}

var tree = new Folder() { Name = "Root" };
var folder1 = new Folder() { Name = "Folder1" };
var file1 = new TreeConverter.File() { Name = "File1" };
var file2 = new TreeConverter.File() { Name = "File2" };
folder1.Children = new List<INode> { file1, file2 };
var folder2 = new Folder() { Name = "Folder2" };
var folder3 = new Folder() { Name = "Folder3" };
var file3 = new TreeConverter.File() { Name = "File3" };
var file4 = new TreeConverter.File() { Name = "File4" };
folder3.Children = new List<INode> { file3 };
folder2.Children = new List<INode>() { folder3, file4 };
tree.Children = new List<INode>() { folder1, folder2 };

List<Folder> folders = ConvertTreeToListofFoldersWithEachEntryNamedRootToLastFolderWithAllContainingFileNodes(tree);
// 返回的folders中每个文件夹的路径名称应为：[Folder1, Folder2.Folder3, Folder2]
// 可通过遍历每个节点的子节点获取对应的文件节点

简化节点结构后的解决方案 #

我后来简化了节点结构，通过统一的Node类结合枚举区分类型，实现了需求，代码如下：

public enum NodeType
{
    Folder,
    File
}
public class Node
{
    public string Name { get; set; }
    public List<Node> Children { get; set;}
    public NodeType Type { get; set; }

    public Node(string name, NodeType type)
    {
        this.Name = name;
        this.Type = type;
        this.Children = new List<Node>();
    }
}

// 构建树结构
var tree = new Node("Root", NodeType.Folder);
var folder1 = new Node("Folder1", NodeType.Folder );
var file1 = new Node ("File1", NodeType.File );
var file2 = new Node ("File2", NodeType.File );
var file3 = new Node ("File3", NodeType.File );
var file4 = new Node ("File4", NodeType.File );
var folder2 = new Node ("Folder2", NodeType.Folder );
var folder3 = new Node ("Folder3", NodeType.Folder );

folder1.Children = new List<Node> { file1, file2 };
folder3.Children = new List<Node> { file3 };
folder2.Children = new List<Node>() { folder3, file4 };
tree.Children = new List<Node>() { folder1, folder2 };


// 执行转换
var foldersWithFiles = GetFoldersWithFiles(tree);

foreach(var folder in foldersWithFiles)
{
    Console.WriteLine($"文件夹: {folder.Key}");
    Console.WriteLine($" 文件: {string.Join(",", folder.Value)}");
}

Dictionary<string, List<string>> GetFoldersWithFiles(Node root)
{
    var foldersWithFiles = new Dictionary<string, List<string>>();
    Traverse(root, root.Name, foldersWithFiles);
    return foldersWithFiles;
}

void Traverse(Node node, string path, Dictionary<string, List<string>> foldersWithFiles)
{
    // 获取当前节点下的所有文件子节点
    var childFiles = node.Children.Where(child => child.Type == NodeType.File);
    if (childFiles.Any())
    {
        foldersWithFiles[path] = childFiles.Select(child => child.Name).ToList();
    }

    // 递归遍历所有子文件夹节点
    foreach (var child in node.Children.Where(child => child.Type == NodeType.Folder))
    {
        Traverse(child, $"{path}/{child.Name}", foldersWithFiles);
    }
}

运行输出 #

文件夹: Root/Folder1
 文件: File1,File2
文件夹: Root/Folder2
 文件: File4
文件夹: Root/Folder2/Folder3
 文件: File3

内容的提问来源于stack exchange，提问作者Heems