平衡二叉搜索树BST转换为双向链表

递归做法：

1. 分别将BST的左、右子树转换成双向链表
2. new出一个链表节点，值等于BST根节点的值
3. 由于是BST，所以new出的节点应该位于链表的中间，所以分别连接左、右子树转换成的链表。这一步中须要找到左链表的尾节点。

对于一些细节处理，要加上必要的判空语句（链表节点为空时，不能访问它）

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } * Definition of Doubly-ListNode * class DoublyListNode { * public: *     int val; *     DoublyListNode *next, *prev; *     DoublyListNode(int val) { *         this->val = val;           this->prev = this->next = NULL; *     } * } */class Solution {public:    /**     * @param root: The root of tree     * @return: the head of doubly list node     */    DoublyListNode* bstToDoublyList(TreeNode* root) {        if (!root) return NULL;                DoublyListNode *left = bstToDoublyList(root->left);        DoublyListNode *right = bstToDoublyList(root->right);                DoublyListNode *left_tail = left;        while (left_tail && left_tail->next) left_tail = left_tail->next;                DoublyListNode *cur = new DoublyListNode(root->val);        cur->prev = left_tail;        if (left_tail) left_tail->next = cur;        cur->next = right;        if (right) right->prev = cur;                if (left) return left;        else return cur;    }};