2265 - Count Nodes Equal to Average of Subtree (Medium)

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.

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int ans = 0;
    pair<int, int> dfs(TreeNode* node) {
        if (!node) return {0, 0}; // {sum, cnt}
        // post order traversal
        auto l = dfs(node->left);
        auto r = dfs(node->right);
        // sum from left tree + sum from right tree + current node value
        int sum = l.first + r.first + node->val;
        // cnt from left tree + cnt from right tree + current node value
        int cnt = l.second + r.second + 1;
        // check if the avgerage is same as the node value
        ans += (sum / cnt == node->val);
        // return the pair
        return {sum, cnt};
    }

    int averageOfSubtree(TreeNode* root) {
        dfs(root);
        return ans;
    }
};