[PROBLEM] 2265. Count Nodes Equal to Average of Subtree

[godkingjay]

Difficulty

Medium

Problem 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:

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

Link

https://leetcode.com/problems/count-nodes-equal-to-average-of-subtree/

[sinhaaayush10]

Count Nodes Equal to Average of Subtree Solution::- class Solution { public: int result;

pair<int, int> solve(TreeNode* root) {
    if(!root)
        return {0, 0};
    
    pair<int, int> l = solve(root->left);
    pair<int, int> r = solve(root->right);
    
    int leftSum   = l.first;
    int leftCount = l.second;
    
    int rightSum   = r.first;
    int rightCount = r.second;
    
    int SUM   = leftSum + rightSum + root->val;
    int COUNT = leftCount + rightCount + 1;
    
    int avg = SUM/COUNT;
    
    if(avg == root->val) {
        result++;
    }
    
    return {SUM, COUNT};
}

int averageOfSubtree(TreeNode* root) {
    result = 0;
    
    solve(root);
    
    return result;
}

};

Approach-Doing postorder traversal