108. 将有序数组转换为二叉搜索树 

 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 public:
13     TreeNode* sortedArrayToBST(vector<int>& nums) 
14     {
15         if(nums.size() == 0) return NULL;
16         int l = 0,r = nums.size() - 1;
17         return slove(nums,l,r);
18     }
19 
20     TreeNode* slove(vector<int>& nums,int l,int r) 
21     {
22         int mid = (l + r)/2;
23         TreeNode* root = new TreeNode(nums[mid]);
24         if(mid != l) root->left = slove(nums,l,mid - 1);
25         else root->left = 0;
26 
27         if(mid != r) root->right = slove(nums,mid + 1,r);
28         else root->right = 0;
29 
30         return root;
31     }
32 };

 

posted @ 2020-04-01 14:16  Jinxiaobo0509  阅读(119)  评论(0)    收藏  举报