The average value of nodes on level 0 is 3, on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11]. Note: The range of node's value is in the range of 32-bit signed integer. 给一个非空二叉树,返回每层的平均值组成的数组。 解法:BFS迭代,层序遍历,计算每层的平均...
nodes= [xfornodeinnodesforxin(node.left, node.right)ifx]returnans 解法2: DFS #Definition for a binary tree node.#class TreeNode(object):#def __init__(self, x):#self.val = x#self.left = None#self.right = NoneclassSolution(object):defaverageOfLevels(self, root):""":type root: T...
LeetCode - 637. Average of Levels in Binary Tree 链接637. Average of Levels in Binary Tree 题意 给定非空二叉树,求出每一层数的平均值 思路 利用队列存储每一层的数,存完之后需要取出size,再循环求平均值。这样保证了循环的次数就是每一层的结点数。 代码......
leetcode专题—637. Average of Levels in Binary Tree 原文:https://leetcode.com/problems/average-of-levels-in-binary-tree/discuss/105107/Java-BFS-Solution 题目意思:求每一层节点值和的平均数。BFS,用队列存每一level的节点,在除以size...Leet...
leetcode Add to List 637. Average of Levels in Binary Tree,就是对二叉树每一层求平均值。开始想法用层次遍历,然而又不知道该点是在哪一层。所以要求没一点的深度,而通过先序遍历递归求没一点的深度的过程中。好像并不需要再去层次遍历,结果就已经很明显了。要注意l
Note: The range of node's value is in the range of 32-bit signed integer. 2.解决思路和代码 循环每层的节点,保存到二维数组,二维数组的一个值保存这一层节点的值之和,另外一个值保存这一层的节点个数。 class Solution: def averageOfLevels(self, root): info = [] self.dfs(info, root, 0...
The range of node's value is in the range of 32-bit signed integer. 思路: 本题具体思路是使用二叉树的层次遍历法,具体可以参考以前写的一篇博客:107.Binary Tree Level Order Traversal II要注意的是,在求平均值时,求和可能会溢出,所以代码中将数据类型定义为long,详情请查看accumulate的相关用法及注意点 ...
Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array. 给定一非空二叉树,返回其每层的平均值所构成的数组。 Example 1: Input: 3 / \ 9 20 / \ 15 7 Output:[3, 14.5, 11] ...
Average of Levels in Binary Tree [思路]:获得树每一层的平均值,使用广度优先搜索。那么使用方法: 广度优先 deque
1#Definition for a binary tree node.2#class TreeNode(object):3#def __init__(self, x):4#self.val = x5#self.left = None6#self.right = None78classSolution(object):9defaverageOfLevels(self, root):10"""11:type root: TreeNode12:rtype: List[float]13"""14levels = [[root]]#将要...