我正在研究樹問題Convert Sorted Array to Binary Search Tree - LeetCode給定一個(gè)元素按升序排序的數(shù)組，將其轉(zhuǎn)換為高度平衡的 BST。對于這個(gè)問題，高度平衡二叉樹被定義為一個(gè)二叉樹，其中每個(gè)節(jié)點(diǎn)的兩個(gè)子樹的深度相差永遠(yuǎn)不會超過 1。例子：Given the sorted array: [-10,-3,0,5,9],One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:      0     / \   -3   9   /   / -10  5直觀的 D&Q 解決方案是class Solution:    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:        """        Runtime: 64 ms, faster than 84.45%        Memory Usage: 15.5 MB, less than 5.70%         """        if len(nums) == 0: return None        #if len(nums) == 1: return TreeNode(nums[0])        mid = len(nums) // 2        root = TreeNode(nums[mid])        if len(nums) == 1: return root         if len(nums) > 1:             root.left = self.sortedArrayToBST(nums[:mid])            root.right = self.sortedArrayToBST(nums[mid+1:])        return root        mid設(shè)置為len(nums)//2或(low + high)//2當(dāng)閱讀其他提交時(shí)，我發(fā)現(xiàn)class Solution:    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:        return self.buildBST(nums, 0, len(nums))    def buildBST(self, nums, left, right):        if right <= left:            return None        if right == left + 1:            return TreeNode(nums[left])        mid = left + (right - left) // 2        root = TreeNode(nums[mid])        root.left = self.buildBST(nums, left, mid)        root.right = self.buildBST(nums, mid + 1, right)        return rootmid 被設(shè)置為 mid = low + (high -low)//2mid = low + (high -low)//2超過有(low + high)//2什么好處？