Count Nodes Equal to Average of Subtree
<p>Given the <code>root</code> of a binary tree, return <em>the number of nodes where the value of the node is equal to the <strong>average</strong> of the values in its <strong>subtree</strong></em>.</p>

<p><strong>Note:</strong></p>

<ul>
	<li>The <strong>average</strong> of <code>n</code> elements is the <strong>sum</strong> of the <code>n</code> elements divided by <code>n</code> and <strong>rounded down</strong> to the nearest integer.</li>
	<li>A <strong>subtree</strong> of <code>root</code> is a tree consisting of <code>root</code> and all of its descendants.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2265.Count%20Nodes%20Equal%20to%20Average%20of%20Subtree/images/image-20220315203925-1.png" style="width: 300px; height: 212px;" />
<pre>
<strong>Input:</strong> root = [4,8,5,0,1,null,6]
<strong>Output:</strong> 5
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2200-2299/2265.Count%20Nodes%20Equal%20to%20Average%20of%20Subtree/images/image-20220326133920-1.png" style="width: 80px; height: 76px;" />
<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> For the node with value 1: The average of its subtree is 1 / 1 = 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>
Tree; Depth-First Search; Binary Tree
/**
 * 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 averageOfSubtree(TreeNode* root) {
        int ans = 0;
        auto dfs = [&](this auto&& dfs, TreeNode* root) -> pair<int, int> {
            if (!root) {
                return {0, 0};
            }
            auto [ls, ln] = dfs(root->left);
            auto [rs, rn] = dfs(root->right);
            int s = ls + rs + root->val;
            int n = ln + rn + 1;
            if (s / n == root->val) {
                ++ans;
            }
            return {s, n};
        };
        dfs(root);
        return ans;
    }
};
