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# Maximum Product of Splitted Binary Tree

Given a binary tree `root`. Split the binary tree into two subtrees by removing 1 edge such that the product of the sums of the subtrees are maximized.

Since the answer may be too large, return it modulo 10^9 + 7.

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/01/21/sample_1_1699.png)

```
Input: root = [1,2,3,4,5,6]
Output: 110
Explanation: Remove the red edge and get 2 binary trees with sum 11 and 10. Their product is 110 (11*10)
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/01/21/sample_2_1699.png)

```
Input: root = [1,null,2,3,4,null,null,5,6]
Output: 90
Explanation:  Remove the red edge and get 2 binary trees with sum 15 and 6.Their product is 90 (15*6)
```

**Example 3:**

```
Input: root = [2,3,9,10,7,8,6,5,4,11,1]
Output: 1025
```

**Example 4:**

```
Input: root = [1,1]
Output: 1
```

**Constraints:**

* Each tree has at most `50000` nodes and at least `2` nodes.
* Each node's value is between `[1, 10000]`.

```java
class Solution {
    long sum, maxProd, mod = 1_000_000_007;

    public void DFS(TreeNode root) {
        if (root == null)
            return;
        sum += (long) root.val;
        DFS(root.left);
        DFS(root.right);
    }

    public long checkMax(TreeNode root) {
        if (root == null)
            return 0L;
        long treeSum = root.val;
        treeSum += checkMax(root.left);
        treeSum += checkMax(root.right);
        maxProd = Math.max(maxProd, treeSum * (sum - treeSum));
        return treeSum;
    }

    public int maxProduct(TreeNode root) {
        sum = maxProd = 0;
        DFS(root);
        long rootProd = checkMax(root);
        return (int) (maxProd % mod);
    }
}
```