108. Convert Sorted Array to Bin

根据已排序数组创建平衡二叉查找树

需要认识到，平衡二叉查找树的中序遍历即为有序数组。

递归解法：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        return self.createBST(nums, 0, len(nums) - 1)
    
    def createBST(self, nums, left, right):
        if left > right:
            return None
        mid = left + (right-left)//2
        print(mid)
        # print(nums[mid])
        root = TreeNode()
        root.val = nums[mid]
        root.left = self.createBST(nums, left, mid-1)
        root.right = self.createBST(nums, mid+1, right)
        return root