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[12/10] 1339. Maximum Product of Splitted Binary Tree
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1339. Maximum Product of Splitted Binary Tree
Given the root of a binary tree, split the binary tree into two subtrees by removing one edge such that the product of the sums of the subtrees is maximized.
Return the maximum product of the sums of the two subtrees. Since the answer may be too large, return it modulo 109 + 7.
Note that you need to maximize the answer before taking the mod and not after taking it.
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)
Constraints:
- The number of nodes in the tree is in the range
[2, 5 * 104]. 1 <= Node.val <= 104
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# 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
from collections import deque
class Solution:
def maxProduct(self, root: Optional[TreeNode]) -> int:
MOD = 10**9+7
queue = deque([root])
allsum = 0
while queue:
node = queue.popleft()
allsum += node.val
if node.left: queue.append(node.left)
if node.right: queue.append(node.right)
max_mult = 0
def dfs(node):
if not node: return 0
nonlocal max_mult
left = dfs(node.left)
right = dfs(node.right)
sub_tree_sum = left+right+node.val
max_mult = max(max_mult, (allsum - sub_tree_sum)*sub_tree_sum)
return sub_tree_sum
dfs(root)
return max_mult%MOD
