# 思路

以 recursive 遍歷將一個升冪的陣列轉成高度平衡的 BST (二元搜尋樹)

# 參考程式碼

	/**
	* 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) {}
	* };
	*/

	static auto fast_io = []
	{
	ios::sync_with_stdio(false);
	cout.tie(nullptr);
	cin.tie(nullptr);
	return 0;
	}();

	class Solution {
	public:
	TreeNode* sortedArrayToBST(vector<int>& nums)
	{
	return BST(0, nums.size() - 1, nums);
	}

	TreeNode* BST(int l, int r, vector<int>& nums)
	{
	if (l > r) return NULL;

	int m = l + (r - l) / 2;
	TreeNode* root = new TreeNode(nums[m]);

	root->left = BST(l, m - 1, nums);
	root->right = BST(m + 1, r, nums);
	return root;
	}
	};