2347. Count Nodes Equal to Average of Subtree

Problem Statement

2347. Count Nodes Equal to Average of Subtree

Medium

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

C++

Source file

/**
 * 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 res = 0;
    pair<int, int> solve(TreeNode* root){
        if(!root)
            return {0, 0};
        pair<int, int> ltSub = solve(root->left);
        pair<int, int> rtSub = solve(root->right);
        int sum = ltSub.first + rtSub.first + root->val;
        int ct = ltSub.second + rtSub.second + 1;
        if(floor(sum/ct)==root->val)  res++;
        return {sum, ct};       
       }

    int averageOfSubtree(TreeNode* root) {
        solve(root);
        return res;
    }
};