题解 | #将升序数组转化为平衡二叉搜索树# | C++

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
class Solution {
private:
    TreeNode* dfs(const vector<int>& nums, int b, int e) {
        if (b > e) {
            return nullptr;
        }
        int mid = b + (e - b)/2 + (e-b)%2;
        TreeNode* node = new TreeNode(nums[mid]);
        node->left = dfs(nums, b, mid-1);
        node->right = dfs(nums, mid+1, e);
        return node;
    }
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* sortedArrayToBST(vector<int>& nums) {
        return dfs(nums, 0, nums.size()-1);
    }
};