题解 | #将升序数组转化为平衡二叉搜索树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param nums int整型vector 
     * @return TreeNode类
     */
    TreeNode*create_BST(vector<int>&nums,int left,int right){
        if(left>right){
            return NULL;
        }
        int mid=(left+right+1)/2;
        TreeNode *root=new TreeNode(nums[mid]);
        root->left=create_BST(nums,left,mid-1);
        root->right=create_BST(nums,mid+1,right);
        return root;

    }

    TreeNode* sortedArrayToBST(vector<int>& nums) {
        // write code here
        if(nums.size()==0){
            return NULL;
        }
        return create_BST(nums,0,nums.size()-1);
    }
};