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.
• 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

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[0] + p2[0] + n->val, count = 1 + p1[1] + p2[1];
    return {sum, count, p1[2] + p2[2] + (n->val == sum / count)};
}
int averageOfSubtree(TreeNode* root) {
    return dfs(root)[2];
}

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[0] + right[0] + node.val;
        int currCount = left[1] + right[1] + 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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