二叉搜索树 | 将有序数组转换为二叉搜索树

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

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

思路:

其实就是构建一个二叉搜索树，对于搜索树，有三个特点:
1、根节点左边的都比根节点小。
2、根节点右边的都比根节点大。
3、左孩子和右孩子都是搜索树。
因此通过递归构建左孩子，右孩子。

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if len(nums) == 1:
            return TreeNode(nums[0])
        elif len(nums) == 0:
            return None
        else:
            mid = (len(nums)-1) / 2
            node = TreeNode(nums[mid])
            node.left = self.sortedArrayToBST(nums[:mid])
            node.right = self.sortedArrayToBST(nums[mid+1:])
            return node