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.

这道题就是个深搜，不过可能我没有减枝还是什么，时间是1188ms,网上的是16ms。但是思想貌似是一样的:

1.当前结点值加上孩子的孩子得curSum。

2.不算当前结点，左子树和右子树相加（包括当前结点的左右孩子）得lrmax;

算出来取最大值返回即可。

不知此题有无非递归法？

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int rob(TreeNode* root) {
if(root==NULL)
return 0;
int leftmax=0;
int rightmax=0;
int cur=root->val;
if(root->left!=NULL){
leftmax=rob(root->left);
cur+=rob(root->left->left)+rob(root->left->right);
}
if(root->right!=NULL){
rightmax=rob(root->right);
cur+=rob(root->right->left)+rob(root->right->right);
}
if(cur<leftmax+rightmax)
cur=leftmax+rightmax;
return cur;
}
};