<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>

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;

const int MOD = 1e9 + 7;

class Solution {
public:
    int maxProduct(TreeNode *root) {
        TreeNode dpRoot;
        dpTree(&dpRoot, root);

        ll ans = 0;
        stack<TreeNode*> s;
        s.push(&dpRoot);
        while(!s.empty()) {
            TreeNode *node = s.top();
            s.pop();
            if(node->left != nullptr) {
                ans = max(ans, (ll)node->left->val * (dpRoot.val - node->left->val));
                s.push(node->left);
            }
            if(node->right != nullptr) {
                ans = max(ans, (ll)node->right->val * (dpRoot.val - node->right->val));
                s.push(node->right);
            }
        }

        return ans % MOD;
    }

    int dpTree(TreeNode *root, TreeNode *node) {
        int left = 0, right = 0;
        if(node->left != nullptr) {
            root->left = new TreeNode();
            left = dpTree(root->left, node->left);
        }
        if(node->right != nullptr) {
            root->right = new TreeNode();
            right = dpTree(root->right, node->right);
        }
        root->val = node->val + left + right;
        return root->val;
    }
};

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