<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://assets.leetcode.com/uploads/2021/04/24/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://assets.leetcode.com/uploads/2021/04/24/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>

/**
 * 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) {
        vector<int> counts;
        vector<int> sums;
        vector<int> res;
        rec(counts, sums, root);
        auto max_el = *max_element(counts.begin(), counts.end());
        cout << max_el<<endl;
        for (int i=0; i<counts.size(); i++) {
            cout << counts[i] << " " << sums[i]<<endl;
            if (counts[i] == max_el) {
                res.push_back(sums[i]);
            }
        }
        return res;
    }
    int rec(vector<int>& counts, vector<int>& sums, TreeNode* node) {
        if (node==nullptr) {
            return 0;
        }
        int sub_sum = rec(counts, sums, node->left)+rec(counts, sums, node->right)+node->val;
        auto find_element = find(sums.begin(), sums.end(), sub_sum);
        if (find_element!=sums.end()) {
            int index=find_element-sums.begin();
            counts[index]++;
        }
        else {
            sums.push_back(sub_sum);
            counts.push_back(1);
        }
        return sub_sum;
    }
};
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