How do you solve Maximum Product of Splitted Binary Tree on LeetCode?

Maximum Product of Splitted Binary Tree (LeetCode #1339) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the relationship between subtree sums and the total sum is crucial for optimizing the solution.

What is the brute force solution for Maximum Product of Splitted Binary Tree?

The brute force approach for Maximum Product of Splitted Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the sum of all possible subtrees by removing each edge one at a time. This allows us to evaluate the product of the sums of the two resulting subtrees for each edge removal. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Product of Splitted Binary Tree?

Maximum Product of Splitted Binary Tree uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search (DFS), Tree Traversal.

What are the interview tips for Maximum Product of Splitted Binary Tree?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as trees with only two nodes, to ensure your solution is robust. Practice explaining your thought process as you code, as this can help you catch mistakes early.

What are common mistakes when solving Maximum Product of Splitted Binary Tree?

Failing to calculate the total sum of the tree before trying to compute products. Not considering the modulo operation until after the product is calculated.

#1339

Maximum Product of Splitted Binary Tree

Medium

TreeDepth-First SearchBinary TreeDepth-First Search (DFS)Tree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach calculates the total sum of the tree once and then uses a single traversal to compute the maximum product by leveraging the relationship between the total sum and subtree sums.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total sum of the binary tree in one traversal.
2Step 2: During the same traversal, calculate the sum of each subtree and compute the product with the remaining tree sum.
3Step 3: Keep track of the maximum product found during the traversal.

solution.py26 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def maxProduct(self, root: TreeNode) -> int:
10        total_sum = self.tree_sum(root)
11        self.max_product = 0
12        self.calculate_product(root, total_sum)
13        return self.max_product % (10**9 + 7)
14
15    def tree_sum(self, node):
16        if not node:
17            return 0
18        return node.val + self.tree_sum(node.left) + self.tree_sum(node.right)
19
20    def calculate_product(self, node, total_sum):
21        if not node:
22            return 0
23        subtree_sum = node.val + self.calculate_product(node.left, total_sum) + self.calculate_product(node.right, total_sum)
24        product = subtree_sum * (total_sum - subtree_sum)
25        self.max_product = max(self.max_product, product)
26        return subtree_sum

ℹ

Complexity note: The time complexity is O(n) because we traverse the tree once to calculate the total sum and again to calculate the maximum product, leading to linear operations.

1Understanding the relationship between subtree sums and the total sum is crucial for optimizing the solution.
2Using a single traversal to gather necessary data can significantly reduce time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.