獲取所有節點在樹上

Question

考慮下面的類，用於構建一個樹層次結構的特定級別：

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

我怎麼能得到一個特定級別的所有對象的列表？

下面是一些示例數據（希望其正確格式化）：

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，使用節點level3childE，我怎麼會從同級別的所有節點？節點名稱可以是任何東西。

Answer 1

這裏是作爲SimpleTest的成員函數實現的（未測試的）遞歸方法：

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; 
    } 
}