> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# House Robber III Python Solution with Tests

> Tested Python solution for LeetCode 337 with 14 pytest cases. Generate a practice environment with lcpy.

LeetCode 337, Medium. Topics: Dynamic Programming, Tree, Depth-First Search, Binary Tree. [View on LeetCode](https://leetcode.com/problems/house-robber-iii/description/).

Generate this problem as a practice environment: tested reference solution, 14 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 337   # by problem number
lcpy gen -s house_robber_iii   # by problem name
```

## 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***.

### Examples

![Example 1](https://assets.leetcode.com/uploads/2021/03/10/rob1-tree.jpg)

```
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](https://assets.leetcode.com/uploads/2021/03/10/rob2-tree.jpg)

```
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

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/house_robber_iii/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/house_robber_iii/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
from leetcode_py import TreeNode


class Solution:
    # Time: O(n)
    # Space: O(h)
    def rob(self, root: TreeNode[int] | None) -> int:
        def dfs(node: TreeNode[int] | None) -> tuple[int, int]:
            # Returns (max if rob node, max if skip node).
            if node is None:
                return 0, 0
            left_rob, left_skip = dfs(node.left)
            right_rob, right_skip = dfs(node.right)
            rob = node.val + left_skip + right_skip
            skip = max(left_rob, left_skip) + max(right_rob, right_skip)
            return rob, skip

        return max(dfs(root))
```

## Complexity

| Time | Space |
| ---- | ----- |
| O(n) | O(h)  |

## Tags

[NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).