Leetcode 1339. Maximum Product of Splitted Binary Tree

2020-02-29 15:44:35 阅读：454 来源： 互联网

标签：node Binary Product 1339 layers subsum None TreeNode root

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].

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担心递归会超时，写了一个非递归的。对于非递归来说，layersorder，parent, subsum这几个是必须的，但是忘了更新layers

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def maxProduct(self, root):
        layerorder = [root]
        parent,subsum = {root:None},{root:0} #bug1: {root,None}
        layers= [[root],[]]
        c,n = 0,1
        while (layers[c]):
            for node in layers[c]:
                for child in [node.left, node.right]:
                    if (child != None):
                        layers[n].append(child) #bug2: miss this line
                        layerorder.append(child)
                        parent[child]=node
                        subsum[child]=0
            layers[c].clear()
            c,n = n,c

        for node in layerorder[::-1]:
            subsum[node] += node.val
            p = parent[node]
            if (p != None):
                subsum[p] += subsum[node]
        total,maxv,maxi = subsum[root],subsum[root]*2,root

        for i in range(1,len(layerorder)):
            diff = abs(subsum[layerorder[i]]*2-total)
            if (diff < maxv):
                maxv = diff
                maxi = layerorder[i]
        return (total-subsum[maxi])*subsum[maxi]%1000000007

s = Solution()
n1 = TreeNode(1)
n2 = TreeNode(2)
n3 = TreeNode(3)
n4 = TreeNode(4)
n5 = TreeNode(5)
n6 = TreeNode(6)
n1.left,n1.right = n2,n3
n2.left,n2.right = n4,n5
n3.left = n6
print(s.maxProduct(n1))

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标签：node,Binary,Product,1339,layers,subsum,None,TreeNode,root
来源： https://blog.csdn.net/taoqick/article/details/104574987

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Leetcode 1339. Maximum Product of Splitted Binary Tree