leetcode: 108. Convert Sorted Array to Binary Search Tree

最新推荐文章于 2022-02-19 22:56:38 发布

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本文介绍了一种将已排序的数组转换为高度平衡的二叉搜索树的方法。平衡二叉搜索树定义为任意节点的两个子树深度差不超过1的二叉树。示例中给出了一个具体实例，通过递归方式选择中间元素作为根节点，然后分别对左右子数组进行相同操作，最终构造出高度平衡的二叉搜索树。

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Difficulty

Easy.

Problem

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

AC

# Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution():
    def sortedArrayToBST(self, nums):
        if not nums:
            return None
        mid = len(nums) // 2
        root = TreeNode(nums[mid])
        root.left = self.sortedArrayToBST(nums[:mid])
        root.right = self.sortedArrayToBST(nums[mid+1:])
        return root