# LeetCode: 337. House Robber III

**The Problem**

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**

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

**Constraints:**

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

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

^{4}

**Solution**

We opted for a recursive approach to tackle this problem, and upon evaluating our solution on the LeetCode platform, we achieved the following outcome:

Here's the code that led us to this result.

```
int rob(TreeNode* root) {
pair<int, int> ans = robHelper(root);
return max(ans.first, ans.second);
}
pair<int, int> robHelper(TreeNode *&root) {
if (root == NULL) return { 0, 0 };
pair<int, int> left = robHelper(root->left);
pair<int, int> right = robHelper(root->right);
int notRobbed =
max( left.first, left.second) +
max(right.first, right.second);
int robbed = root->val + left.first + right.first;
return pair<int, int>{notRobbed, robbed};
}
```

Here's a simplified explanation:

1. `int rob(TreeNode* root)`

: This is the main function that initiates the calculation. It calls the `robHelper`

function to perform the actual computation. It returns the maximum amount of money that can be robbed from the binary tree.

2. `pair<int, int> robHelper(TreeNode *&root)`

: This helper function calculates the maximum amount that can be robbed from the current subtree rooted at `root`

. It returns a pair of values: `{notRobbed, robbed}`

.

- If the
`root`

is`NULL`

, it means there are no houses to rob in this subtree, so it returns`{0, 0}`

. - It recursively calls
`robHelper`

for the left and right child nodes and gets their results as`left`

and`right`

. - It calculates two values:
`notRobbed`

and`robbed`

. `notRobbed`

is the maximum value when the current house is not robbed. It is calculated by taking the maximum of the sum of not robbing the left child and not robbing the right child.`robbed`

is the maximum value when the current house is robbed. It is calculated by adding the value of the current house, the`notRobbed`

value of the left child, and the`notRobbed`

value of the right child.- It returns the pair
`{notRobbed, robbed}`

for the current subtree.