## Question

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 / \ 2 3 \ \ 3 1

Maximum amount of money the thief can rob = `4`

+ `5`

= **9**.

## Solution

**Result:** Accepted **Time:** 15 ms

Here should be some explanations.

```
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
struct State{
int a,b;
int get_max()
{
return a > b ? a : b;
}
State():a(0),b(0){}
State(int aa,int bb):a(aa),b(bb){}
};
State robber(TreeNode * root)
{
if(root == NULL)
return State();
State left = robber(root->left);
State right = robber(root->right);
return State(left.get_max() + right.get_max(), left.a + right.a + root->val);
}
class Solution {
public:
int rob(TreeNode* root) {
return robber(root).get_max();
}
};
```

**Complexity Analytics**

- Time Complexity:
- Space Complexity: