leetcode 337:House Robber III

leetcode更新题目，又是一个入室盗窃。。。。

题目如下：

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.

跟入室盗窃系列1和2差不多的思想，动态规划即可。然而，这个是树，不能像以前一样，一个vector+遍历搞定（事实上，不需要一个数组，两个int即可，一个记录前一个最大值，一个计记录前前个最大值）。

怎么办呢？思想一样，记录当前节点最大值以及当前节点下一层的最大值。因为是树，得递归，而且一次返回两个参数，不妨用结构体。用结构体保存两个参数，层层递归即可。

结构体定义如下：

typedef struct maxMoney{
int now;//全局最优，即包含根节点的最优
int pre;//不含根节点最优，即根节点下一层左子树和右子数的全局最优之和
}MAXMONEY;

CPP可AC代码如下：

class Solution {
public:
    typedef struct maxMoney{
int now;//全局最优
int pre;//不含根节点最优
}MAXMONEY;
    int max(int a,int b){
        return a>b?a:b;
    }
    MAXMONEY robmoney(TreeNode* root){
        MAXMONEY money;
        money.now=0;
money.pre=0;
        if (root==NULL)//空节点则返回值均为0
            return money;
MAXMONEY moneyleft=robmoney(root->left);//递归求左子树
MAXMONEY moneyright=robmoney(root->right);//递归求右子树

money.now=max((root->val+moneyleft.pre+moneyright.pre),(moneyleft.now+moneyright.now));//全局最优
money.pre=moneyleft.now+moneyright.now;//不含根节点的最优
        return money;
    }
    int rob(TreeNode* root) {
MAXMONEY money=robmoney(root);

        return money.now;
    }
};