题解 | #将升序数组转化为平衡二叉搜索树#

注意中点的取值，以及终止条件

class Solution {
public:
 /**
  *
  * @param num int整型vector
  * @return TreeNode类
  */
 TreeNode *sortedArrayToBST(vector<int> &num) {
     if (num.size() < 1) return nullptr;
     TreeNode *root = preOrder(num, 0, num.size() - 1);
     return root;
 }

 TreeNode *preOrder(vector<int> &num, int left, int right) {
     if (left > right) return nullptr;//注意是，左大于右
     // 1. 找到中间节点
     int mid = left + (right - left + 1) / 2;//注意这个中点的取值方法
     // 2. 递归构建左右
     TreeNode *root = new TreeNode(num[mid]);
     root->left = preOrder(num, left, mid - 1);
     root->right = preOrder(num, mid + 1, right);
     return root;
 }
};