Most Frequent Subtree Sum
<p>Given the <code>root</code> of a binary tree, return the most frequent <strong>subtree sum</strong>. If there is a tie, return all the values with the highest frequency in any order.</p>

<p>The <strong>subtree sum</strong> of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0508.Most%20Frequent%20Subtree%20Sum/images/freq1-tree.jpg" style="width: 207px; height: 183px;" />
<pre>
<strong>Input:</strong> root = [5,2,-3]
<strong>Output:</strong> [2,-3,4]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0508.Most%20Frequent%20Subtree%20Sum/images/freq2-tree.jpg" style="width: 207px; height: 183px;" />
<pre>
<strong>Input:</strong> root = [5,2,-5]
<strong>Output:</strong> [2]
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>
Tree; Depth-First Search; Hash Table; 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:
    vector<int> findFrequentTreeSum(TreeNode* root) {
        unordered_map<int, int> cnt;
        int mx = 0;
        function<int(TreeNode*)> dfs = [&](TreeNode* root) -> int {
            if (!root) {
                return 0;
            }
            int s = root->val + dfs(root->left) + dfs(root->right);
            mx = max(mx, ++cnt[s]);
            return s;
        };
        dfs(root);
        vector<int> ans;
        for (const auto& [k, v] : cnt) {
            if (v == mx) {
                ans.push_back(k);
            }
        }
        return ans;
    }
};
