1339. Maximum Product of Splitted Binary Tree
https://leetcode.com/problems/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:
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
50000nodes and at least2nodes. - Each node's value is between
[1, 10000].
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Intuition
Compute total of all node values
At each node - find sum of subtree rooted at that node
Calculate product of sum(subtree) * (total - sum(subtree))
Track max product as long ans
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Time - O(N)
Space - O(N) - word case for skewed tree
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| /** | |
| * Definition for a binary tree node. | |
| * public class TreeNode { | |
| * int val; | |
| * TreeNode left; | |
| * TreeNode right; | |
| * TreeNode() {} | |
| * TreeNode(int val) { this.val = val; } | |
| * TreeNode(int val, TreeNode left, TreeNode right) { | |
| * this.val = val; | |
| * this.left = left; | |
| * this.right = right; | |
| * } | |
| * } | |
| */ | |
| class Solution { | |
| int total; | |
| // product can be long before % MOD | |
| long ans; | |
| int MOD = (int) 1e9 + 7; | |
| public int maxProduct(TreeNode root) { | |
| if (root == null) { | |
| return 0; | |
| } | |
| total = 0; | |
| ans = 0; | |
| getTotal(root); | |
| dfs(root); | |
| return (int) (ans % MOD); | |
| } | |
| private int dfs(TreeNode node) { | |
| if (node == null) { | |
| return 0; | |
| } | |
| int sum = node.val + dfs(node.left) + dfs(node.right); | |
| // total - current sub tree .. cast as long to avoid int overflow | |
| ans = Math.max(ans, (long)(total - sum) * sum); | |
| return sum; | |
| } | |
| private void getTotal(TreeNode node) { | |
| if (node == null) { | |
| return; | |
| } | |
| total += node.val; | |
| getTotal(node.left); | |
| getTotal(node.right); | |
| } | |
| } |
