[leetcode: Python] 108. Convert Sorted Array to Binary Search Tree

最新推荐文章于 2025-02-03 15:37:10 发布

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本文介绍两种将有序数组转换为高度平衡的二叉搜索树的方法。第一种方法递归地选取数组中间元素作为根节点，并继续处理左右子数组。第二种方法采用辅助函数，通过传入左右边界实现更高效的递归构建。

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题目：
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

方法一：性能92ms

# 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
        """
        length = len(nums)
        if length == 0:
            return None
        if length == 1:
            return TreeNode(nums[0])
        root = TreeNode(nums[length / 2])
        root.left = self.sortedArrayToBST(nums[:length / 2])
        root.right = self.sortedArrayToBST(nums[length / 2 + 1:])
        return root

方法二：性能82ms

# 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 
        return self.helper(nums, 0, len(nums) - 1)

    def helper(self, nums, left, right):
        if left > right:
            return
        mid = (left + right) // 2
        node = TreeNode(nums[mid])
        left = self.helper(nums, left, mid - 1)
        right = self.helper(nums, mid + 1, right)
        node.left = left 
        node.right = right
        return node