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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.

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

Java

  • /**
     * 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<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            while (!queue.isEmpty()) {
                TreeNode node = queue.poll();
                nodesList.add(node);
                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;
        }
    }
    
  • // 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
    
    
    

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