# [LintCode]378.将二叉查找树转换成双链表

``````    4
/ \
2   5
/ \
1   3
``````

``````/**
* 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 == NULL)
return NULL;

vector<int> list;
treeToList(root, list);//先把值存到数组

}
void treeToList(TreeNode *TreeRoot, vector<int> &list) {
//树结点的值通过中序遍历存入数组list
if(TreeRoot != NULL) {
treeToList(TreeRoot->left, list);
list.push_back(TreeRoot->val);
treeToList(TreeRoot->right, list);
}
}
//通过已知数组构造双链表
DoublyListNode* listToDlist(vector<int> &list) {
head = p = new DoublyListNode(list[0]);//数组首位赋值
for(int i=1; i<list.size(); i++) {     //从数组第二位开始搜索
DoublyListNode* q = new DoublyListNode(list[i]);
p->next = q;
q->prev = p;
p = q;//将q插入到链表尾部
}