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Formatted question description: https://leetcode.ca/all/1339.html

# 1339. 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].

Related Topics:
Dynamic Programming, Tree

## Solution 1.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxProduct(TreeNode root) {
final int MODULO = 1000000007;
List<TreeNode> nodesList = new ArrayList<TreeNode>();
Map<TreeNode, TreeNode> childParentMap = new HashMap<TreeNode, TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
TreeNode left = node.left, right = node.right;
if (left != null) {
childParentMap.put(left, node);
queue.offer(left);
}
if (right != null) {
childParentMap.put(right, node);
queue.offer(right);
}
}
for (int i = nodesList.size() - 1; i > 0; i--) {
TreeNode node = nodesList.get(i);
TreeNode parent = childParentMap.get(node);
if (parent != null)
parent.val += node.val;
}
long maxProduct = 0;
for (int i = nodesList.size() - 1; i > 0; i--) {
TreeNode node = nodesList.get(i);
int value1 = node.val, value2 = root.val - node.val;
long product = (long) value1 * (long) value2;
maxProduct = Math.max(maxProduct, product);
}
int maxProductModulo = (int) (maxProduct % MODULO);
return maxProductModulo;
}
}

############

/**
* 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 {
private long ans;
private long s;
private static final int MOD = (int) 1e9 + 7;

public int maxProduct(TreeNode root) {
s = sum(root);
dfs(root);
ans %= MOD;
return (int) ans;
}

private long sum(TreeNode root) {
if (root == null) {
return 0;
}
return root.val + sum(root.left) + sum(root.right);
}

private long dfs(TreeNode root) {
if (root == null) {
return 0;
}
long t = root.val + dfs(root.left) + dfs(root.right);
if (t < s) {
ans = Math.max(ans, t * (s - t));
}
return t;
}
}

• // OJ: https://leetcode.com/problems/maximum-product-of-splitted-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
long total = 0, ans = 0, mod = 1e9 + 7;
void sum(TreeNode *root) {
if (!root) return;
total += root->val;
sum(root->left);
sum(root->right);
}
long dfs(TreeNode *root) {
if (!root) return 0;
long s = root->val + dfs(root->left) + dfs(root->right);
ans = max(ans, (total - s) * s);
return s;
}
public:
int maxProduct(TreeNode* root) {
sum(root);
dfs(root);
return ans % mod;
}
};

• # 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:
def sum(root):
if root is None:
return 0
return root.val + sum(root.left) + sum(root.right)

def dfs(root):
nonlocal s, ans
if root is None:
return 0
t = root.val + dfs(root.left) + dfs(root.right)
if t < s:
ans = max(ans, t * (s - t))
return t

s = sum(root)
ans = 0
dfs(root)
ans %= 10**9 + 7
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func maxProduct(root *TreeNode) int {
mod := int(1e9) + 7
var sum func(*TreeNode) int
sum = func(root *TreeNode) int {
if root == nil {
return 0
}
return root.Val + sum(root.Left) + sum(root.Right)
}
s := sum(root)
ans := 0
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
t := root.Val + dfs(root.Left) + dfs(root.Right)
if t < s {
ans = max(ans, t*(s-t))
}
return t
}
dfs(root)
return ans % mod
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function maxProduct(root: TreeNode | null): number {
const sum = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
return root.val + sum(root.left) + sum(root.right);
};
const s = sum(root);
let ans = 0;
const mod = 1e9 + 7;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
const t = root.val + dfs(root.left) + dfs(root.right);
if (t < s) {
ans = Math.max(ans, t * (s - t));
}
return t;
};
dfs(root);
return ans % mod;
}