【leetcode】108. Convert Sorted Array to Binary Search Tree

题目如下：

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

解题思路：我的方法是取nums的中间元素作为根节点，nums的左半部分为左子树，右半部分为右子树，同时继续对左右子树分别递归。

代码如下：

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

class Solution(object):
    def recursive(self,node,nums):
        mid = len(nums)/2
        left_num = nums[:mid]
        if len(left_num) > 0:
            node.left = TreeNode(left_num[len(left_num)/2])
            self.recursive(node.left,left_num)
        right_num = nums[mid+1:]
        if len(right_num) > 0:
            node.right = TreeNode(right_num[len(right_num)/2])
            self.recursive(node.right,right_num)

    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if len(nums) == 0:
            return None
        root = TreeNode(nums[len(nums)/2])
        self.recursive(root,nums)
        return root

原文地址：https://www.cnblogs.com/seyjs/p/10625297.html