<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 dfs(self, root, vis):
        if root==None:
            return 0

        vis[root]=1
        # print (root.val)
        l_sum = 0
        if root.left and root.left not in vis:
            l_sum = self.dfs(root.left, vis)
        r_sum = 0
        if root.right and root.right not in vis:
            r_sum = self.dfs(root.right, vis)
        
        return l_sum+r_sum+root.val

    def dfs1(self, root, vis, tot, ans):
        if root==None:
            return 0

        vis[root]=1
        l_sum = 0
        if root.left and root.left not in vis:
            l_sum = self.dfs1(root.left, vis, tot, ans)
        r_sum = 0
        if root.right and root.right not in vis:
            r_sum = self.dfs1(root.right, vis, tot, ans)
        
        ans[0] = max(ans[0], max(l_sum*(tot-l_sum), r_sum*(tot-r_sum)))
        
        
        return l_sum+r_sum+root.val
        

        

    def maxProduct(self, root: Optional[TreeNode]) -> int:
        vis = {}
        tot = self.dfs(root, vis)
        # print (tot)
        ans = [0]
        vis = {}
        self.dfs1(root, vis, tot, ans)
        return ans[0]%(1000000000+7)
199