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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<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;
        }
    }
    
    ############
    
    /**
     * 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;
    }
    
    

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