99. 恢复二叉搜索树 

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution 
11 {
12     vector<int> num;
13     int i = 0;
14     void inorder(TreeNode* root)
15     {
16         if(!root) return;
17         inorder(root->left);
18         num.push_back(root -> val);
19         inorder(root->right);
20     }
21 
22     void recover(TreeNode* root)
23     {
24         if(!root) return;
25         recover(root -> left);
26         root -> val = num[i++];
27         recover(root -> right);
28     }
29 public:
30     void recoverTree(TreeNode* root) 
31     {
32         inorder(root);
33         sort(num.begin(),num.end());
34         recover(root);
35     }
36 };

 

posted @ 2020-05-06 11:29  Jinxiaobo0509  阅读(110)  评论(0)    收藏  举报