<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:

        # first get the sum of the tree.
        # then transversal the tree, and consider cut the current node, get its sum. then we can get the product of all 

        global total 
        total = 0 

        def transversal(root): 
            global total
            if root == None: 
                return 
            total += root.val 
            transversal(root.left)
            transversal(root.right)

        transversal(root)

        global ans
        ans = 0

        def transversal2(root): 
            global total, ans
            if root == None: 
                return 0 
            
            res1 = transversal2(root.left)
            res2 = transversal2(root.right)
            res = res1 + res2 + root.val
            ans = max(ans, res * (total - res))
            return res 
        
        transversal2(root)
        return ans %(10**9 + 7)

        
        
171