1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 /**
10  * Definition for a binary tree node.
11  * struct TreeNode {
12  *     int val;
13  *     TreeNode *left;
14  *     TreeNode *right;
15  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
16  * };
17  */
18 class Solution {
19 public:
20     TreeNode* sortedListToBST(ListNode* head) {
21         if(!head)
22             return NULL;
23         if(head->next==NULL){
24             TreeNode*root=new TreeNode(head->val);
25             return root;
26         }
27         ListNode*pre=getLeftOfMid(head),*mid=pre->next;
28         TreeNode*root=new TreeNode(mid->val);//根
29         pre->next=NULL;
30         root->left=sortedListToBST(head);//左子树
31         root->right=sortedListToBST(mid->next);//右子树
32         return root;
33     }
34     ListNode* getLeftOfMid(ListNode*head)//说简单点，就是获取链表中间节点，作为树的根节点，并由此划分根的2个子树
35     {
36         if(!head)
37             return NULL;
38         ListNode*pre=head,*back=head;
39         while(back!=NULL)//back后退的步数大致是head的两倍，确保pre能落在中间节点位置的前一结点处
40         {
41             back=back->next;
42             if(!back)
43                 break;
44             back=back->next;
45             if(!back)
46                break;
47             pre=head;
48             head=head->next;
49         }
50         return pre;
51     }
52 };

 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 private:
12      int cur;//不去掉这前面的static编译不会通过的
13 public:
14     TreeNode* sortedArrayToBST(vector<int>& nums) {
15         cur = 0;
16         if (nums.size() == 0)
17             return NULL;
18         return helper(nums, 0, nums.size() - 1);
19     }
20     TreeNode* helper(vector<int>&nums, int l, int r)
21     {
22         if (l > r)
23             return NULL;
24         int mid = (l+r) / 2;
25         TreeNode*left=helper(nums, l, mid-1);
26         TreeNode * root = new TreeNode(nums[cur++]);//回忆考研时候，中序遍历中，这个地方才是真正干活的语句（并且终须遍历二叉排序树得到的就是有序数组）
27         root->left = left;
28         root->right = helper(nums, mid + 1, r);
29         return root;
30     }
31 };