2021-12-08 637. 二叉树的层平均值【Easy】

给定一个非空二叉树, 返回一个由每层节点平均值组成的数组。

示例 1：

输入：
    3
   / \
  9  20
    /  \
   15   7
输出：[3, 14.5, 11]
解释：
第 0 层的平均值是 3 ,  第1层是 14.5 , 第2层是 11 。因此返回 [3, 14.5, 11] 。

提示：

节点值的范围在32位有符号整数范围内。

方法一：DFS

# 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 == None:
            return []
        
        sum_count = []

        ans = []
        preorder(root, 0, sum_count)
        for item in sum_count:
            ans.append(item[0]/item[1])
        return ans

def preorder(root, depth, sum_count):
    if root == None: return;
    if depth >= len(sum_count): sum_count.append([0, 0])
    sum_count[depth][0] += root.val
    sum_count[depth][1] += 1
    preorder(root.left, depth+1, sum_count)
    preorder(root.right, depth+1, sum_count)

方法二：BFS

# 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 == None:
            return []
        
        ans = []

        curr = []
        next = []
        curr.append(root)

        while len(curr) != 0:
            sum = 0
            for node in curr:
                sum += node.val
                if node.left: next.append(node.left)
                if node.right: next.append(node.right)
            ans.append(sum/len(curr))
            curr = next
            next = []
        return ans