把二元查找树转变成排序的双向链表

1.把二元查找树转变成排序的双向链表
题目：
输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表。
要求不能创建任何新的结点，只调整指针的指向。

10
/ /
6 14
/ / / /
4 8 12 16

转换成双向链表
4=6=8=10=12=14=16。

C# codes as below:

using System;

namespace ConsoleApp

{

class RunClass

{

static void Main()

{

Node<int> i1 = new Node<int>() { Value = 4 };

Node<int> i2 = new Node<int>() { Value = 8 };

Node<int> i3 = new Node<int>() { Value = 12 };

Node<int> i4 = new Node<int>() { Value = 16 };

Node<int> i5 = new Node<int>() { Value = 6, Left = i1, Right = i2 };

Node<int> i6 = new Node<int>() { Value = 14, Left = i3, Right = i4 };

Node<int> i7 = new Node<int>() { Value = 10, Left = i5, Right = i6 };

Tree<int> myTree = new Tree<int>() { Root = i7 };

Link<int> myLink = new Helper<int>().Transfer(myTree);

Node<int> current = myLink.Start;

Console.WriteLine(current.Value);

while (current.Right != null)

{

current = current.Right;

Console.WriteLine(current.Value);

}

Console.ReadKey();

}

}

class Helper<T>

{

public Link<T> Transfer(Tree<T> tree)

{

Node<T> root = tree.Root;

if (root.Left != null)

{

Link<T> leftLink = Transfer(new Tree<T>() { Root = root.Left });

Node<T> leftNode = leftLink.Start;

while (leftNode.Right != null)

{

leftNode = leftNode.Right;

}

root.Left = leftNode;

leftNode.Right = root;

}

if (root.Right != null)

{

Link<T> rightLink = Transfer(new Tree<T>() { Root = root.Right });

Node<T> rightNode = rightLink.Start;

root.Right = rightNode;

rightNode.Left = root;

}

Node<T> start=root;

while(start.Left!=null)

{

start=start.Left;

}

return new Link<T>() { Start = start };

}

}

class Node<T>

{

public Node<T> Left { get; set; }

public Node<T> Right { get; set; }

public T Value { get; set; }

}

class Tree<T>

{

public Node<T> Root { get; set; }

}

class Link<T>

{

public Node<T> Start { get; set; }

}

}