如何递归地从孩子 c# 中获取所有父母？

Question

有所有视图位置（比如说 100，这些是 TreeView 的位置）：Id、Name、ParentId 1 Root Null 2 Semi-root 1 3 Semi-semi-root 2 4.....

还有一些数据，我们只从另一个请求中收到与我们的值相对应的位置：Id、Name、ParentId 22 Location1 12 36 Location38 21 99 Location38 3

需要获取我们从请求中获得的数据的所有父层次结构。 有 class：

public class TreeViewNode {
  public Guid Id {get; set;}
  public string Name {get; set;}
  public Guid ParentId {get; set;}
}

测试数据：

 private IEnumerable<TreeViewNode> SeedData()
        {
            return new List<TreeViewNode>
            {
                new()
                {
                    Id = 1,
                    Name = "Root",
                    ParentId = null
                },
                new()
                {
                    Id = 2,
                    Name = "Semi-root",
                    ParentId = 1
                },
                new()
                {
                    Id = 3,
                    Name = "Semi te",
                    ParentId = 2
                },
                new()
                {
                    Id = 4,
                    Name = "Semi oi",
                    ParentId = 2
                },
                new()
                {
                    Id = 5,
                    Name = "Child",
                    ParentId = 3
                },
                new()
                {
                    Id = 6,
                    Name = "Child 1",
                    ParentId = 4
                },
                new()
                {
                    Id = 7,
                    Name = "Child 2",
                    ParentId = 1
                },
                new()
                {
                    Id = 8,
                    Name = "Child 3",
                    ParentId = 1
                },
                new()
                {
                    Id = 9,
                    Name = "Child 4",
                    ParentId = 1
                },
                new()
                {
                    Id = 10,
                    Name = "Child 6",
                    ParentId = 2
                }
            };
        }

从请求中获得的示例数据：

 var dataFromRequest = new List<TreeViewNode>
            {
                new()
                {
                    Id = 8,
                    Name = "Child 3",
                    ParentId = 1
                },
                new()
                {
                    Id = 10,
                    Name = "Child 6",
                    ParentId = 2
                },
                new()
                {
                    Id = 33,
                    Name = "Child",
                    ParentId = 3
                },
                new()
                {
                    Id = 4,
                    Name = "Semi oi",
                    ParentId = 2
                }
            };

结果需要得到这样的父母名单：

Id Name ParentId
1 Root Null
2 Semi-root 1
3 Semi te 2

这里可以是5个或更多级别的父母

Answer 1

像这样的东西

// create Dictionary for lookups
var lookup = dataFromRequest.ToDictionary(x => x.Id.Value, x => x);

var lookFor = new TreeViewNode()
{
    Id = 6,
    Name = "Child 1",
    ParentId = 4
};

// get all the parents
GetParents(lookup, lookFor);

获取所有父母的帮助方法

private void GetParents(Dictionary<int, TreeViewNode> lookup, 
                                TreeViewNode lookFor)
{
    while (lookFor != null)
    {
        // Alternative: Add lookFor to List<TreeViewNode> here and return at 
        // the end of the method
        Debug.WriteLine($"{lookFor.Id} {lookFor.Name} {lookFor.ParentId}");

        if (lookFor.ParentId == null)
            break;

        // cast ParentId to corrent dataType here Guid or int
        lookup.TryGetValue((int)lookFor.ParentId, out var parentNode);
        lookFor = parentNode;
    }

}

当您获得更多IEnumerable<TreeViewNode>时，请记住让您的查找字典保持最新

您可以做的另一件事是在TreeViewNode中设置父节点，如下所示：

public class TreeViewNode {
    ...
    // set ParentNode by lookup in the Dictionary once
    public TreeViewNode ParentNode {get; set;}
}