网站 301b2b多平台一键发布

描述

给定一个非空二叉树的根节点 root , 以数组的形式返回每一层节点的平均值。与实际答案相差 10-5 以内的答案可以被接受

思路

无需思路，乱杀

解答

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:if root is None:return []all_level = [root, ]ret = []while all_level:level_cnt = 0level_total = 0nxt_level = []for node in all_level:if node:level_cnt+=1level_total+=node.valnode.left and nxt_level.append(node.left) or Nonenode.right and nxt_level.append(node.right) or Noneif level_cnt > 0:ret.append(level_total/level_cnt)else:ret.append(None)all_level = nxt_levelreturn ret