<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:
    int sum(TreeNode* root){
        //ajamos yvelaperi
        if(!root) return 0;
        int ls=sum(root->left);
        int rs=sum(root->right);
        return root->val +ls +rs;
    }
    void helper(TreeNode* root, vector<int>& sums){
        if(!root) return;
        sums.push_back(sum(root));
        helper(root->left,sums);
        helper(root->right,sums);
    }
    vector<int> findFrequentTreeSum(TreeNode* root) {
        if(!root) return {};
        vector<int> sums;
        helper(root,sums);
        //sums has all the sums of the tree
        //make a map and find the most frequent member
        map<int,int> mp;
        for(int i=0;i<sums.size();i++){
            mp[sums[i]]++;
        }
        vector<int> ans;
        //add most frequent elements
        int mf=0;
        for(auto it:mp){
            mf=max(mf,it.second);
        }
        //mf is the frequency which is the highest
        
        for(auto i:mp){
            if(i.second==mf){
                ans.push_back(i.first);
            }
        }


        return ans;
    }
};
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