【力扣hot100题】（041）将有序数组转换为二叉搜索树

从中间开始，将左右两边的数构建成子树，递归即可。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        int len=nums.size();
        if(len==0) return nullptr;
        vector<int> numsleft(nums.begin(),nums.begin()+len/2);
        vector<int> numsright(nums.begin()+len/2+1,nums.end());
        TreeNode* root=new TreeNode(nums[len/2]);
        root->left=sortedArrayToBST(numsleft);
        root->right=sortedArrayToBST(numsright);
        return root;
    }
};