【LeetCode】637. Average of Levels in Binary Tree 解答・解説【Python】

問題
1. 原文
2. 内容
解答
1. 解答1：Python, BFS(幅優先探索)
2. 解答2：
メモ・参考・感想

問題

原文

Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

Input: root = [3,9,20,null,null,15,7] Output: [3.00000,14.50000,11.00000] Explanation: 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].

1
2
3
4

Input: root = [3,9,20,null,null,15,7]
Output: [3.00000,14.50000,11.00000]
Explanation: 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].

Example 2:

Input: root = [3,9,20,15,7] Output: [3.00000,14.50000,11.00000]

1
2

Input: root = [3,9,20,15,7]
Output: [3.00000,14.50000,11.00000]

Constraints:

The number of nodes in the tree is in the range [1, 10⁴].

-2³¹ <= Node.val <= 2³¹ - 1

内容

二分木rootが与えられるので、各階層の値の平均値を配列で返してください。

誤差は10^(-5)以内としてください。

※正しくない可能性があります。

解答

解答1：Python, BFS(幅優先探索)

# 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 not root:
            return None
        
        answer = []
        q = deque()
        q.append(root)
        
        while q:
            sum_of_value = 0
            length_of_q = len(q)
            for node in range(len(q)):
                node = q.popleft()
                if node.val:sum_of_value += node.val                
                if node.left: q.append(node.left)
                if node.right: q.append(node.right)
            answer.append(sum_of_value/length_of_q)
        
        return answer

# 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 not root:

return None

answer = []

q = deque()

q.append(root)

while q:

sum_of_value = 0

length_of_q = len(q)

for node in range(len(q)):

node = q.popleft()

if node.val:sum_of_value += node.val

if node.left: q.append(node.left)

if node.right: q.append(node.right)

answer.append(sum_of_value/length_of_q)

return answer

幅優先探索

各階層の探索を行う際、valの合計値をsum_of_valueに、各階層のノードの数をlength_of_qにそれぞれ保存しておき、各階層の探索終了時に平均値を求めてanswerに追加する。

解答2：

メモ・参考・感想

前：107. Binary Tree Level Order Traversal II

次：202. Happy Number

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