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Problem solution · Python

Maximum Product of Splitted Binary Tree

Maximum Product of Splitted Binary Tree: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
29 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Maximum Product of Splitted Binary Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 29 lines of Python from the credited upstream file maximum-product-of-splitted-binary-tree.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Product of Splitted Binary Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(h) # Definition for a binary tree node.class TreeNode(object):    def __init__(self, x):        self.val = x        self.left = None        self.right = None  class Solution(object):    def maxProduct(self, root):        """        :type root: TreeNode        :rtype: int        """        MOD = 10**9 + 7        def dfs(root, total, result):            if not root:                return 0            subtotal = dfs(root.left, total, result)+dfs(root.right, total, result)+root.val            result[0] = max(result[0], subtotal*(total-subtotal) )            return subtotal         result = [0]        dfs(root, dfs(root, 0, result), result)        return result[0] % MOD

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