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# Count Nodes Equal to Average of Subtree LeetCode Solution

## Problem – Count Nodes Equal to Average of Subtree leetCode Solution

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.
• 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 = 
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`

### Count Nodes Equal to Average of Subtree LeetCode Solution in C++

``````array<int, 3> dfs(TreeNode* n) {
if (n == nullptr)
return {0, 0, 0};
auto p1 = dfs(n->left), p2 = dfs(n->right);
int sum = p1 + p2 + n->val, count = 1 + p1 + p2;
return {sum, count, p1 + p2 + (n->val == sum / count)};
}
int averageOfSubtree(TreeNode* root) {
return dfs(root);
}
``````

### Count Nodes Equal to Average of Subtree LeetCode Solution in Java

``````class Solution {
int res = 0;
public int averageOfSubtree(TreeNode root) {
dfs(root);
return res;
}

private int[] dfs(TreeNode node) {
if(node == null) {
return new int[] {0,0};
}

int[] left = dfs(node.left);
int[] right = dfs(node.right);

int currSum = left + right + node.val;
int currCount = left + right + 1;

if(currSum / currCount == node.val) {
res++;
}

return new int[] {currSum, currCount};
}
}
``````

### Count Nodes Equal to Average of Subtree LeetCode Solution in python

``````class Solution:
def averageOfSubtree(self, root: Optional[TreeNode]) -> int:

def calculate_average(root):
if root:
self.summ+=root.val
self.nodecount+=1
calculate_average(root.left)
calculate_average(root.right)

def calculate_for_each_node(root):
if root:
self.summ = 0
self.nodecount = 0
calculate_average(root)
if ((self.summ)//(self.nodecount)) == root.val:
self.count+=1
calculate_for_each_node(root.left)
calculate_for_each_node(root.right)

self.count = 0
calculate_for_each_node(root)
return self.count

``````
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