2265. Count Nodes Equal to Average of Subtree

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2265. Count Nodes Equal to Average of Subtree

Description

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:

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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:

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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

Hints/Notes

Solution

Language: C++

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/**
 * 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;

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

    pair<int, int> dfs(TreeNode* root) {
        if (!root) {
            return {0, 0};
        }
        auto left = dfs(root->left);
        auto right = dfs(root->right);
        int totalSum = left.first + right.first + root->val;
        int totalNode = left.second + right.second + 1;
        if (totalSum / totalNode == root->val) {
            res++;
        }
        return {totalSum, totalNode};
    }
};