<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 solve(TreeNode* root, vector<int> &ans) {
    if (root == NULL) {
        return 0;
    }
    int leftSum = solve(root->left, ans);
    int rightSum = solve(root->right, ans);
    int sum = leftSum + rightSum + root->val;
    ans.push_back(sum); // Push the sum of this subtree into the vector
    return sum;
}

    vector<int> findFrequentTreeSum(TreeNode* root) {
        vector<int>ans;
        solve(root,ans);
        map<int,int>mp;

        for(auto i:ans){
            mp[i]++;
        }
        vector<pair<int,int>>pairr;

        for(auto i:mp){
            pairr.push_back({i.second,i.first});
        }
        int maxi = INT_MIN;
        for(auto i:pairr){
            maxi = max(maxi,i.first);
        }
        vector<int>finalAns;
        for(auto i:pairr){
            if(i.first==maxi){
                finalAns.push_back(i.second);
            }
        }
        return finalAns;
    }
};
23000