<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 {
unordered_map<int,int>mp;

public:

    int findSum(TreeNode* root) {
        if(!root) return 0;

        int left = findSum(root->left);
        int right = findSum(root->right);

        int sum = left+right+root->val;
        mp[sum]++;

        return sum;
    }


    vector<int> findFrequentTreeSum(TreeNode* root) {
        
        vector<int>ans;
        if(!root) return ans;
        priority_queue<pair<int,int>>pq;
        findSum(root);
        for(auto it=mp.begin(); it!=mp.end();it++)
        pq.push({it->second,it->first});

        int maxx=pq.top().first;

        while(!pq.empty() && pq.top().first==maxx) {
            ans.push_back(pq.top().second);
            pq.pop();
        }

        return ans;

        
    }
};
24600