二叉搜索树与双向链表

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。

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     std::vector<TreeNode*> vv;
13     void inorder(TreeNode* root)
14     {
15         if (root ->left != NULL)
16         {
17             inorder(root->left);
18         }
19         vv.push_back(root);
20         if(root->right != NULL)
21         {
22             inorder(root->right);
23         }
24     }
25     TreeNode* Convert(TreeNode* pRootOfTree)
26     {
27         if (pRootOfTree == NULL)
28             return NULL;
29         inorder(pRootOfTree);
30 
31         if(vv.size() == 1)
32             return pRootOfTree;
33 
34         vv[0]->left = NULL;
35         vv[0]->right = vv[1];
36         int i = 1;
37         for (i; i <= vv.size() - 2; ++i)
38         {
39             vv[i]->left = vv[i-1];
40             vv[i]->right = vv[i+1];
41         }
42         vv[i]->left = vv[i-1];
43         vv[i]->right = NULL;
44         return vv[0];
45     }
46 };