337. House Robber III 【M】【vip】

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本文介绍了一种使用树形动态规划方法解决特定抢劫问题的算法。在一个特殊的区域中，所有房屋构成了一个二叉树结构，且相邻房屋被抢会导致报警。文章详细解释了如何通过递归方式求解最大抢劫金额，并提供了Python实现代码。

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

Example 1:

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7 .

Example 2:

Maximum amount of money the thief can rob = 4 + 5 = 9 .

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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本质上是一个树的遍历问题

这个刚开始想错了，为什么不能用广搜呢，因为并不是如果我不rob树根，就一定要rob他的两个子节点。也就是说，有可能我只rob他的一个子节点会更好。

f1(node) be the value of maximum money we can rob from the subtree with node as root ( we can rob node if necessary).

f2(node) be the value of maximum money we can rob from the subtree with node as root but without robbing node.

Then we have

f2(node) = f1(node.left) + f1(node.right) and

f1(node) = max( f2(node.left)+f2(node.right)+node.value, f2(node) ).

class Solution(object):
    def do(self,root):
        if root == None:
            return (0,0)

        l = self.do(root.left)
        r = self.do(root.right)

        return (l[1] + r[1] , max(l[1] + r[1] , l[0] + r[0] + root.val))

    def rob(self, root):
        return self.do(root)[1]