Maximum Product of Splitted Binary Tree

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Three simple concepts in this problem, here is the problem: https://leetcode.com/problems/maximum-product-of-splitted-binary-tree/

1343. Maximum Product of Splitted Binary Tree
Medium
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:
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:
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].
It can be solved in O(n)-time and O(1)-space, using the following concepts:

  1. Utilization of Post-Order Traversal (SetValSumNodes)
  2. Reusing the node val to store the sum of all the elements for that sub-tree
  3. BigInteger computation (Numerics library) followed by the mod 10^9 + 7
Notice that due to #2, you can compute the current product in constant time (current product: (root.val - currentNode.left.val) * currentNode.left.val). Code is below, cheers, ACC.


public class Solution
{
    public int MaxProduct(TreeNode root)
    {
        if (root == null) return 0;

        BigInteger val = -1;
        SetValSumNodes(root);
        MaxProduct(root, root.val, ref val);
        return (int)(val % 1000000007);
    }

    private void MaxProduct(TreeNode node,
                            int sumAllNodes,
                            ref BigInteger prod)
    {
        if (node == null || (node.left == null && node.right == null)) return;
        if (node.left != null)
        {
            BigInteger current = (BigInteger)(sumAllNodes - node.left.val) * node.left.val;
            prod = BigInteger.Max(prod, current);
        }
        if (node.right != null)
        {
            BigInteger current = (BigInteger)(sumAllNodes - node.right.val) * node.right.val;
            prod = BigInteger.Max(prod, current);
        }
        MaxProduct(node.left, sumAllNodes, ref prod);
        MaxProduct(node.right, sumAllNodes, ref prod);
    }

    private void SetValSumNodes(TreeNode node)
    {
        if (node == null) return;
        SetValSumNodes(node.left);
        SetValSumNodes(node.right);
        if (node.left != null) node.val += node.left.val;
        if (node.right != null) node.val += node.right.val;
    }
}

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