题目描述 输入一棵二叉搜索树，将该二叉搜索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，只能调整树中结点指针的指向。

 private TreeNode2 lca = null;
    public TreeNode2 lowestCommonAncestor(TreeNode2 root, TreeNode2 p, TreeNode2 q) {
        if (root == null){
            return null;
        }
        findNode(root,p,q);
        return lca;
    }

    //看从root出发能不能找到p和q
    private boolean findNode(TreeNode2 root, TreeNode2 p, TreeNode2 q) {
        if (root == null){
            return false;
        }
        int left = findNode(root.left,p,q)?1:0;
        int right = findNode(root.right,p,q)?1:0;
        int mid = (root == p || root == q)?1:0;
        if (left + right + mid == 2){
            lca = root;
        }
        return (left + right + mid)>0;
    }

    //输入一棵二叉搜索树，将该二叉搜索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，只能调整树中结点指针的指向。
    //中序遍历
    public TreeNode2 Convert(TreeNode2 pRootOfTree) {
        if (pRootOfTree == null){
            return null;
        }
        if (pRootOfTree.left == null && pRootOfTree.right == null){
            return pRootOfTree;
        }

        //最终的链表 =》 左子树链表 + 根节点 + 右子树的链表
        //1.先递归处理左子树
        //left就是左子树这个链表的根节点
        TreeNode2 left = Convert(pRootOfTree.left);
        //2.需要找到左子树链表的尾节点
        //right相当于链表的next
        TreeNode2 leftTail = left;
        while (leftTail != null && leftTail.right != null){
            leftTail = leftTail.right;
        }
        //循环结束之后，leftTail就指向了左侧链表的尾部
        //3.把左子树和当前节点连在一起
        if (left != null){
            leftTail.right = pRootOfTree;
            pRootOfTree.left = leftTail;
        }
        //4.递归转换右子树，把右子树也变成双向链表
        TreeNode2 right = Convert(pRootOfTree.right);
        //5.把当前节点和右子树连在一起
        if (right != null){
            right.left = pRootOfTree;
            pRootOfTree.right = right;
        }
        //6.最终返回新的链表的头节点
        return left == null ? pRootOfTree : left;
    }