[Leetcode]Convert Sorted Array to Binary Search Tree

最新推荐文章于 2022-02-19 22:56:38 发布

原创最新推荐文章于 2022-02-19 22:56:38 发布 · 154 阅读

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介绍如何将一个按升序排列的数组转换为高度平衡的二叉搜索树（BST）。平衡二叉搜索树定义为对于每一个节点，其两个子树的深度之差不超过1。示例中给出了一种可能的解决方案。

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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

递归：

# 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 sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if not nums:
            return None
        medium = len(nums)//2
        node = TreeNode(nums[medium])
        node.left = self.sortedArrayToBST(nums[:medium])
        node.right = self.sortedArrayToBST(nums[medium+1:])
        
        return node