108.将有序数组转换为二叉搜索树

思路：

• 因为要创建平衡的二叉搜索树，所以利用二分查找和递归 定义辅助函数help，left是左边下标，high是右边下标。
• 进入while循环，循环条件为left<=right,定义mid为left和right的中点，创建根节点root，值等于nums[mid].
• 开始递归，将左子树和右子树拼接到root上，最后返回root

/**
 * 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 n=nums.size();
    return help(nums,0,n-1);
    }

    TreeNode* help(vector<int> nums,int left,int right){
        while(left<=right){
        int mid=left+(right-left)/2;

        TreeNode*root=new TreeNode(nums[mid]);
        root->left=help(nums,left,mid-1);
        root->right=help(nums,mid+1,right);

        return root;
        }
      return nullptr;
    }
};