leetcode1343

Posted 2021-03-12 AsenYang

 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         if root.left == None and root.right == None:
15             self.memo[root] = root.val
16             return root.val
17         elif root.left != None and root.right != None:
18             cur,left,right = 0,0,0
19             if root in self.memo:
20                 return self.memo[root]
21             else:
22                 cur = root.val
23             if root.left in self.memo:
24                 left = self.memo[root.left]
25             else:
26                 left = self.subTreeSum(root.left)
27             if root.right in self.memo:
28                 right = self.memo[root.right]
29             else:
30                 right = self.subTreeSum(root.right)
31             self.memo[root] = cur + left + right
32             return cur + left + right
33         elif root.left != None:
34             cur,left,right = 0,0,0
35             if root in self.memo:
36                 return self.memo[root]
37             else:
38                 cur = root.val
39             if root.left in self.memo:
40                 left = self.memo[root.left]
41             else:
42                 left = self.subTreeSum(root.left)
43             self.memo[root] = cur + left + right
44             return cur + left
45         elif root.right != None:
46             cur,left,right = 0,0,0
47             if root in self.memo:
48                 return self.memo[root]
49             else:
50                 cur = root.val
51             if root.right in self.memo:
52                 right = self.memo[root.right]
53             else:
54                 right = self.subTreeSum(root.right)
55             self.memo[root] = cur + left + right
56             return cur + right
57 
58     def maxProduct(self, root: ‘TreeNode‘) -> int:
59         self.preOrder(root)
60         #print(self.sums)
61         #print(self.subTrees)
62         result = 0
63         for i in range(len(self.subTrees)):
64             part1 = self.subTrees[i]
65             part2 = self.sums - part1
66             product = part1 * part2
67             result = max(result,product)
68         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。