还是单链表排序的问题（C#）

这个问题是一个叫wartim的仁兄提供的代码，很方便。谢谢

namespace WindowsFormsApplication34
{
public partial class Form1 : Form
{
public class LinkedList
{
public LinkedList node = null;
public int data;
}

class Tree
{
public Tree Left = null;
public Tree Right = null;
public int Value;
}

LinkedList LL, Head;
Tree T = null;

public Form1()
{
InitializeComponent();
LinkedList Temp = null;

// 给 LL 赋点值。。。
LL = new LinkedList();
LL.data = 10;
Temp = LL;
LL = new LinkedList();
LL.data = 34;
LL.node = Temp;
Temp = LL;
LL = new LinkedList();
LL.data = 6;
LL.node = Temp;
Temp = LL;
LL = new LinkedList();
LL.data = 52;
LL.node = Temp;
Temp = LL;
LL = new LinkedList();
LL.data = 6;
LL.node = Temp;
Temp = LL;
LL = new LinkedList();
LL.data = 7;
LL.node = Temp;
Head = LL;

// 构建排序二叉树
for (LL = Head; LL != null; LL = LL.node)
AddToTree(ref T, LL.data);

// 输出排序结果
String S = String.Empty;
GetResult(ref T, ref S);
MessageBox.Show(S);
}

private void Form1_Load(object sender, EventArgs e)
{

}

void AddToTree(ref Tree T, int value)
{
if (T == null)
{
T = new Tree();
T.Value = value;
}
else if (value < T.Value) // 比当前值小于的往左
AddToTree(ref T.Left, value);
else if (value >= T.Value) // 比当前值大于等于的往右
AddToTree(ref T.Right, value);
}

void GetResult(ref Tree T, ref String S)
{
if (T == null)
return; // 已经没有节点了
if (T.Left != null)
GetResult(ref T.Left, ref S); // 向左走
S += T.Value.ToString() + " ";
if (T.Right != null)
GetResult(ref T.Right, ref S); // 向右走
}
}
}