2265. Count Nodes Equal to Average of Subtree

2265. 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

Companies: Amazon, Facebook

Related Topics:
Tree, Depth-First Search, Binary Tree

Similar Questions:

• Maximum Average Subtree (Medium)
• Insufficient Nodes in Root to Leaf Paths (Medium)
• Count Nodes Equal to Sum of Descendants (Medium)

Solution 1. DFS

// OJ: https://leetcode.com/problems/count-nodes-equal-to-average-of-subtree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    pair<int, int> dfs(TreeNode *root) {
        if (!root) return {0,0};
        auto [lc, ls] = dfs(root->left);
        auto [rc, rs] = dfs(root->right);
        ans += root->val == (root->val + ls + rs) / (1 + lc + rc);
        return {lc + rc + 1, ls + rs + root->val};
    }
public:
    int averageOfSubtree(TreeNode* root) {
        dfs(root);
        return ans;
    }
};