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

将一个二叉查找树按照中序遍历转换成双向链表。

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样例

给定一个二叉查找树：

    4
   / \
  2   5
 / \
1   3

返回 1<->2<->3<->4<->5。

标签 

/**
 * 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) {
        // write your code here
        if(!root)
            return NULL;
        DoublyListNode* result=Order(root);
        while(result->prev)
            result=result->prev;
        return result;
        
    }
    
    DoublyListNode* Order(TreeNode * root)
    {
        if(!root)
            return NULL;
        
        DoublyListNode* node=new DoublyListNode(root->val);
        
        if(root->left)
        {
            DoublyListNode* left=Order(root->left);
            while(left->next) left=left->next;
            node->prev=left;
            left->next=node;
        }
            
        if(root->right)
        {
            DoublyListNode* right=Order(root->right);
            while(right->prev) right=right->prev;
            node->next=right;
            right->prev=node;
        }    
        
        return node;
    }
};