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.

這道題就是個深搜，不過可能我沒有減枝還是什麼，時間是1188ms,網上的是16ms。但是思想貌似是一樣的:

1.當前結點值加上孩子的孩子得curSum。

2.不算當前結點，左子樹和右子樹相加（包括當前結點的左右孩子）得lrmax;

算出來取最大值返回即可。

不知此題有無非遞歸法？

/**
 * 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;
    }
};