标签:一个 which more eve node ret 节点 param sort
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.
给定一个升序的矩阵,将它转换成一棵平衡二叉树。
其中,平衡二叉树是指树中每个节点子树高度差的绝对值不超过1的树。
思路:由于左右高度差的绝对值不能超过1,所以树根节点应选择升序列表中间的元素,即nums[len(nums)//2]。
该元素前面的元素(比根节点小的元素)同理递归地放入左子树;后面的元素递归地放入右子树。
最后返回各个节点。
代码
class Solution:
def sortedArrayToBST(self, nums):
"""
@param type nums: List[int]
@param rtype: TreeNode
"""
if nums == None:
return
if len(nums) == 1:
return nums[0]
root = nums[len(nums)//2]
self.sortedArrayToBST(nums[:len(nums)//2])
self.sortedArrayToBST(nums[len(nums)//2+1:])
return root
108. Convert Sorted Array to Binary Search [Python]
标签:一个 which more eve node ret 节点 param sort
原文地址:https://www.cnblogs.com/wyz-2020/p/12386366.html