[Leetcode/파이썬]637. Average of Levels in Binary Tree

Average of Levels in Binary Tree

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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].

Example 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

 

# 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]:
        q = deque()
        q.append(root)
        averages = []

        while q :
            level_sum, num_nodes = 0, len(q)
            
            for _ in range(num_nodes) :
                node = q.popleft()
                level_sum += node.val
                if node.left :
                    q.append(node.left)
                if node.right :
                    q.append(node.right)

            averages.append(level_sum / num_nodes)

        return averages

TreeNode{val: 3, 
left: TreeNode{val: 9, left: None, right: None}, 
right: TreeNode{val: 20, left: TreeNode{val: 15, left: None, right: None}, 
			right: TreeNode{val: 7, left: None, right: None}}}

print(root) 를 해보면 이런 트리노드를 확인할 수 있을 것이다. 이런 자료형임을 확인하였다면 deque에 root를 넣어 가장 위에 있는 node를 뽑아온다. 

node.val = 3

node.left.val = 9
node.right.val=20

node.right.left.val = 15
node.right.right.val = 7

이 되므로, [3, 29/2, 22/2] 가 되므로, [3, 14.5, 11] 이 된다. 

리트코드는 이런 문제가 제법 있어서 연습 열심히 해야겠다.