leetcode-108 将有序数组转化为二叉搜索树

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

---

解题要点：

1.以为已经有序，就取中间数值作为根节点，然后层层地推下去就行

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if len(nums) == 0:
            return None
        if len(nums) == 1:
            return TreeNode(nums[0])
        mid = int(n/2)
        root = TreeNode(nums[mid])
        l1 = nums[:mid]
        l2 = nums[mid+1:]
        root.left = self.sortedArrayToBST(l1)
        root.right = self.sortedArrayToBST(l2)
        return root