[leetcode] 337. House Robber III

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:

     3
    / \
   2   3
    \   \ 
     3   1

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

Example 2:

     3
    / \
   4   5
  / \   \ 
 1   3   1

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

今天的题目和198题（传送门）、213题（传送门）相关，大家可以先看下这两题。题目难度为Medium。

和上面提到的两题不同，那两道题用到了动态规划，而这道题其实是深度优先遍历二叉树。拿root（第0层）为例，如果取第0层的节点，则第1层的节点不能取；如果不取第0层的节点，则第1层的节点可取可不取。这样我们需要记录下每个节点取与不取时能够获取的最大钱数，通过深度优先遍历二叉树，最后取root节点返回的两个数值中大的就可以了。具体代码：

class Solution {
    vector getMoney(TreeNode* node) {
        vector ret(2, 0);
        if(!node) return ret;
        vector lRet = getMoney(node->left);
        vector rRet = getMoney(node->right);
        ret[0] = lRet[1] + rRet[1] + node->val;
        ret[1] = max(lRet[0], lRet[1]) + max(rRet[0], rRet[1]);
        return ret;
    }
public:
    int rob(TreeNode* root) {
        vector ret = getMoney(root);
        return max(ret[0], ret[1]);
    }
};