0

给定以下类，用于构建树层次结构：

public class simpletest
{
    public simpletest Parent { get; set; }
    public IList<simpletest> Children { get; set; }
    public string Name { get; set; }
}

如何获取特定级别的所有对象的列表？

这是一些示例数据（希望其格式正确）：

var root = new simpletest() { Name = "Root" };
var level1childA = new simpletest() { Name = "level1childA", Parent = root };
var level1childB = new simpletest() { Name = "level1childB", Parent = root };

var level2childA = new simpletest() { Name = "level2childA", Parent = level1childA };
var level2childB = new simpletest() { Name = "level2childB", Parent = level1childA };
var level2childC = new simpletest() { Name = "level2childC", Parent = level1childB };
var level2childD = new simpletest() { Name = "level3childD", Parent = level1childB };

var level3childA = new simpletest() { Name = "level3childA", Parent = level2childA };
var level3childB = new simpletest() { Name = "level3childB", Parent = level2childA };
var level3childC = new simpletest() { Name = "level3childC", Parent = level2childB };
var level3childD = new simpletest() { Name = "level3childD", Parent = level2childB };
var level3childE = new simpletest() { Name = "level3childE", Parent = level2childC };
var level3childF = new simpletest() { Name = "level3childF", Parent = level2childC };
var level3childG = new simpletest() { Name = "level3childG", Parent = level3childD };
var level3childH = new simpletest() { Name = "level3childH", Parent = level3childD };

level2childA.Children = new List<simpletest> { level3childA, level3childB };
level2childB.Children = new List<simpletest> { level3childC, level3childD };
level2childC.Children = new List<simpletest> { level3childE, level3childF };
level2childD.Children = new List<simpletest> { level3childG, level3childH };

level1childA.Children = new List<simpletest> { level2childA, level2childB };
level1childB.Children = new List<simpletest> { level2childC, level2childD };

因此，例如，使用节点 level3childE，我将如何从同一级别获取所有节点？节点名称可以是任何名称。

4

2 回答 2

1

这是作为 simpletest 的成员函数实现的（未经测试的）递归方法： 

IEnumerable<simpletest> ElementsAtDepth(int depth) {
    if(depth > 0) {
        foreach(simpletest child in this.Children)
            foreach(simpletest element in child.ElementsAtDepth(depth - 1))
                yield return element;
    }
    else {
        foreach(simpletest element in this.Children)
            yield return element;
    }
}
于 2012-08-05T13:47:41.650 回答
0

BlackBear 的解决方案返回下一级并且不能返回可能对绑定有用的根/初始节点。下面的代码将为您提供所请求级别的节点。另请注意，深度与初始节点相关，而不是实际树。

public IEnumerable<simpletest> NodesAtDepth(int depth)
{
    if (depth <= 0 )
        yield return this;
    else if (depth > 1)
    {
        if (this.Children != null)
            foreach (simpletest child in this.Children)
                foreach (simpletest element in child.NodesAtDepth(depth - 1))
                    yield return element;
    }
    else
    {
        if (this.Children != null)
            foreach (simpletest element in this.Children)
                yield return element;
    }
}
于 2018-05-07T01:29:13.247 回答