1 class Solution: 2 def __init__(self): 3 self.subTrees = [] 4 self.sums = 0 5 self.memo = {} 6 def preOrder(self,root): 7 if root != None: 8 self.sums += root.val 9 self.subTrees.append(self.subTreeSum(root)) 10 self.preOrder(root.left) 11 self.preOrder(root.right) 12 13 def subTreeSum(self,root): 14 cur,left,right = 0,0,0 15 if root != None and root in self.memo: 16 return self.memo[root] 17 else: 18 cur = root.val if root != None else 0 19 if root != None and root.left != None and root.left in self.memo: 20 left = self.memo[root.left] 21 else: 22 left = self.subTreeSum(root.left) if root != None and root.left != None else 0 23 if root != None and root.right != None and root.right in self.memo: 24 right = self.memo[root.right] 25 else: 26 right = self.subTreeSum(root.right) if root != None and root.right != None else 0 27 cur = cur + left + right 28 self.memo[root] = cur 29 return cur 30 31 def maxProduct(self, root: 'TreeNode') -> int: 32 self.preOrder(root) 33 #print(self.sums) 34 #print(self.subTrees) 35 result = 0 36 for i in range(len(self.subTrees)): 37 part1 = self.subTrees[i] 38 part2 = self.sums - part1 39 product = part1 * part2 40 result = max(result,product) 41 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。
