leetcode1339

class Solution:
    def __init__(self):
        self.subTrees = []
        self.sums = 0
        self.memo = {}
    def preOrder(self,root):
        if root != None:
            self.sums += root.val
            self.subTrees.append(self.subTreeSum(root))
            self.preOrder(root.left)
            self.preOrder(root.right)

    def subTreeSum(self,root):
        cur,left,right = 0,0,0
        if root != None and root in self.memo:
            return self.memo[root]
        else:
            cur = root.val if root != None else 0
        if root != None and root.left != None and root.left in self.memo:
            left = self.memo[root.left]
        else:
            left = self.subTreeSum(root.left) if root != None and root.left != None else 0
        if root != None and root.right != None and root.right in self.memo:
            right = self.memo[root.right]
        else:
            right = self.subTreeSum(root.right) if root != None and root.right != None else 0
        cur = cur + left + right
        self.memo[root] = cur
        return cur

    def maxProduct(self, root: 'TreeNode') -> int:
        self.preOrder(root)
        #print(self.sums)
        #print(self.subTrees)
        result = 0
        for i in range(len(self.subTrees)):
            part1 = self.subTrees[i]
            part2 = self.sums - part1
            product = part1 * part2
            result = max(result,product)
        return result % (1000000007)

算法思路：二叉树遍历+memo缓存。

遍历二叉树(本题使用先序遍历)，同时计算每一棵子树的所有节点的和。使用缓存记录已经计算过的子树和，以加快算法的执行效率。

得到所有子树的和之后，用树所有节点的和减去任意一个子树的和，将原二叉树分为两部分。

计算这两部分的乘积，保留乘积的最大值。最终结果对1000000007取余，即为所求。