2265. Count Nodes Equal to Average of Subtree

제목

문제)

2265. Count Nodes Equal to Average of Subtree

 

Given the root of a binary tree, return the number of nodes where the value of the node is equal to the average of the values in its subtree.

Note:

• The average of n elements is the sum of the n elements divided by n and rounded down to the nearest integer.
• A subtree of root is a tree consisting of root and all of its descendants.

 

Example 1:

Input: root = [4,8,5,0,1,null,6]
Output: 5
Explanation: 
For the node with value 4: The average of its subtree is (4 + 8 + 5 + 0 + 1 + 6) / 6 = 24 / 6 = 4.
For the node with value 5: The average of its subtree is (5 + 6) / 2 = 11 / 2 = 5.
For the node with value 0: The average of its subtree is 0 / 1 = 0.
For the node with value 1: The average of its subtree is 1 / 1 = 1.
For the node with value 6: The average of its subtree is 6 / 1 = 6.

Example 2:

Input: root = [1]
Output: 1
Explanation: For the node with value 1: The average of its subtree is 1 / 1 = 1.

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• 0 <= Node.val <= 1000

솔루션1)

- postorder 재귀를 사용한다.

- postorder 재귀 함수는 자신 포함 하위 서브트리의 sum 값과 자식 수를 리턴한다.

# 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 averageOfSubtree(self, root: Optional[TreeNode]) -> int:        
        count = 0
        def postorder(node):
            nonlocal count 
            
            if node is None:
                return 0, 0
            
            l_val, l_child_count = postorder(node.left)
            r_val, r_child_count = postorder(node.right)
            
            sum_val = node.val + l_val + r_val
            node_count = 1 + l_child_count + r_child_count
            average = int(sum_val / node_count)
            
            if average == node.val:
                count += 1
                
            return sum_val, node_count
        
        postorder(root)
        return count