二叉查找树变成有序的双向列表

通过中序遍历可以得到一个有序的序列，不创建一个节点实现有序双向列表。

 /// <summary>
    /// 二叉查找树变成有序的双向链表，不创建新的节点
    /// </summary>
    public class BST2LinkList
    {
        public static void Convert()
        {
            Node[] tree = CreateBST();

            Node head = null;
            MidddleOrderSearch(tree, 0, ref head);
            Console.Read();
        }

        public static void MidddleOrderSearch(Node[] tree, int cur, ref Node head)
        {
            //左子结点
            int left = (cur + 1)*2 - 1;
            if (left < tree.Length )
            {
                MidddleOrderSearch(tree, left, ref head); //递归遍历左子树
            }

            //处理当前节点
            if (head == null)
            {
                head = tree[cur];
            }
            else
            {
                AppendNode(head, tree[cur]);
            }

            //右子结点
            int right = (cur + 1) * 2;
            if (right < tree.Length )
            {
                MidddleOrderSearch(tree, right,ref head);//递归遍历右子树
            }
        }

        public static void AppendNode(Node head, Node node)
        {
            Node tmp = head;
            while (tmp.Next != null)
            {
                tmp = tmp.Next;
            }
            tmp.Next = node;
            node.Pre = tmp;
        }

        public static void DisplayLinkList(Node head)
        {
            do
            {
                Console.Write(head.Num + ",");
                head = head.Next;
            }
            while (head != null);
            Console.WriteLine();
        }

        public static void DisplayLinkListFromTail(Node tail)
        {
            while (tail != null&&tail.Next!=null) { tail = tail.Next; };
            do
            {
                Console.Write(tail.Num + ",");
                tail = tail.Pre;
            }
            while (tail != null);
            Console.WriteLine();
        }

        public static Node[] CreateBST()
        {
            Node[] tree=new Node[9];
            int[] arrayNum = { 6,4,8,2,5,7,9,1,3};
            for (int i = 0; i < arrayNum.Length; i++)
            {
                Node tmp = new Node() { Num=arrayNum[i]};
                tree[i] = tmp;
            }
            return tree;
        }

        public class Node
        {
            public int Num;
            public Node Pre;
            public Node Next;
        }
    }