<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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxProduct(self, root: Optional[TreeNode]) -> int:
        sums = []

        def recurse(node):
            if node is None:
                return 0

            if node.right is None and node.left is None:
                sums.append(node.val)
                return node.val
            
            ans = recurse(node.left) + recurse(node.right) + node.val
            sums.append(ans)
            return ans

        max_sum = recurse(root)
            
        return int(max([(max_sum - x) * x for x in sums]) % (math.pow(10, 9) + 7))

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