<p>Given the <code>root</code> of a binary tree, split the binary tree into two subtrees by removing one edge such that the product of the sums of the subtrees is maximized.</p>

<p>Return <em>the maximum product of the sums of the two subtrees</em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p><strong>Note</strong> that you need to maximize the answer before taking the mod and not after taking it.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/21/sample_1_1699.png" style="width: 500px; height: 167px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6]
<strong>Output:</strong> 110
<strong>Explanation:</strong> Remove the red edge and get 2 binary trees with sum 11 and 10. Their product is 110 (11*10)
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/21/sample_2_1699.png" style="width: 500px; height: 211px;" />
<pre>
<strong>Input:</strong> root = [1,null,2,3,4,null,null,5,6]
<strong>Output:</strong> 90
<strong>Explanation:</strong> Remove the red edge and get 2 binary trees with sum 15 and 6.Their product is 90 (15*6)
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 5 * 10<sup>4</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>4</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:

//  long long int dfs(TreeNode*root,vector<long long int>&sum){
//     if(root==NULL)
//     return 0;

//     if(root->left==NULL && root->right==NULL){
//         sum.push_back(root->val);
//          return root->val;

//     }

//   long long   int left=dfs(root->left,sum);
//   long long   int right=dfs(root->right,sum);

//     sum.push_back(left+right+root->val);

//     return left+right+root->val;

   
// }

//     int maxProduct(TreeNode* root) {

//         vector<long long int>sum;

//       long long int totalSum=  dfs(root,sum);

//       long long  int maxi=0;

        

//         for(int i=0;i<sum.size();i++){
//           long long   int first=totalSum-(sum[i]);

//             maxi=max(maxi,(first)*sum[i]);
            
//         }

//         return maxi;

        
//     }
// };



/**
 * 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:
    long long int dfs(TreeNode* root, vector<long long int>& sum) {
        if (root == nullptr)
            return 0;

        if (root->left == nullptr && root->right == nullptr) {
            sum.push_back(root->val);
            return root->val;
        }

        long long int left = dfs(root->left, sum);
        long long int right = dfs(root->right, sum);

        long long int total = left + right + root->val;
        sum.push_back(total);

        return total;
    }

    int maxProduct(TreeNode* root) {
        vector<long long int> sum;
        long long int totalSum = dfs(root, sum);
        long long int maxi = 0;

        for (long long int subtreeSum : sum) {
            long long int complement = totalSum - subtreeSum;
            maxi = max(maxi, (complement) * subtreeSum);
        }

        return maxi % 1000000007; // To handle large numbers, take modulo 10^9+7 as per problem constraints.
    }
};

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