LintCode | 将二叉查找树转换成双向链表

http://www.lintcode.com/zh-cn/problem/convert-binary-search-tree-to-doubly-linked-list/#

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;
    }
};