<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>

class Solution {
private:
int find(TreeNode* root , unordered_map< int , int >&cnt){
    if(root == NULL){
        return 0;
    }
    int left = find(root->left , cnt);
    int right = find(root->right , cnt);
    int sum = root->val + left + right;
    cnt[sum]++;
    return sum; 
}
public:
    vector<int> findFrequentTreeSum(TreeNode* root) {
        unordered_map< int , int >cnt;
        find(root , cnt);
        int value = INT_MIN;
        for( auto i : cnt){
            if(i.second > value){
                value = i.second;
            }
        }
        vector<int>ans;
        for(auto i : cnt){
            if(i.second == value){
                ans.push_back(i.first);
            }
        }
        return ans;
    }
};
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