**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `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 form a binary tree. It will automatically contact the police if **two directly-linked houses were broken into on the same night**.

Given the `root`

of the binary tree, return *the maximum amount of money the thief can rob without alerting the police*.

**Example 1:**

**Input:** root = [3,2,3,null,3,null,1]
**Output:** 7
**Explanation:** Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

**Example 2:**

**Input:** root = [3,4,5,1,3,null,1]
**Output:** 9
**Explanation:** Maximum amount of money the thief can rob = 4 + 5 = 9.

**Constraints:**

- The number of nodes in the tree is in the range
`[1, 10`

.^{4}] `0 <= Node.val <= 10`

^{4}

```
/**
* 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 tryRob(TreeNode* root, int& l, int& r) {
if (!root)
return 0;
int ll = 0, lr = 0, rl = 0, rr = 0;
l = tryRob(root->left, ll, lr);
r = tryRob(root->right, rl, rr);
return max(root->val + ll + lr + rl + rr, l + r);
}
int rob(TreeNode* root) {
int l, r;
return tryRob(root, l, r);
}
};
```

```
public class Solution {
public int rob(TreeNode root) {
int[] num = dfs(root);
return Math.max(num[0], num[1]);
}
private int[] dfs(TreeNode x) {
if (x == null) return new int[2];
int[] left = dfs(x.left);
int[] right = dfs(x.right);
int[] res = new int[2];
res[0] = left[1] + right[1] + x.val;
res[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
return res;
}
}
```

```
class Solution(object):
def rob(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def superrob(node):
# returns tuple of size two (now, later)
# now: max money earned if input node is robbed
# later: max money earned if input node is not robbed
# base case
if not node: return (0, 0)
# get values
left, right = superrob(node.left), superrob(node.right)
# rob now
now = node.val + left[1] + right[1]
# rob later
later = max(left) + max(right)
return (now, later)
return max(superrob(root))
```

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