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Abhishek Chaudhary

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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`

SOLUTION:

``````# 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 sumcount(self, root):
total = 0
n = 0
if root:
ltotal, ln = self.sumcount(root.left)
rtotal, rn = self.sumcount(root.right)
total = ltotal + rtotal + root.val
n = ln + rn + 1
if root.val == total // n:
self.ctr += 1
return (total, n)

def averageOfSubtree(self, root: Optional[TreeNode]) -> int:
self.ctr = 0
self.sumcount(root)
return self.ctr
``````