House Robber III
<p>The thief has found himself a new place for his thievery again. There is only one entrance to this area, called <code>root</code>.</p>

<p>Besides the <code>root</code>, 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 <strong>two directly-linked houses were broken into on the same night</strong>.</p>

<p>Given the <code>root</code> of the binary tree, return <em>the maximum amount of money the thief can rob <strong>without alerting the police</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0337.House%20Robber%20III/images/rob1-tree.jpg" style="width: 277px; height: 293px;" />
<pre>
<strong>Input:</strong> root = [3,2,3,null,3,null,1]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0337.House%20Robber%20III/images/rob2-tree.jpg" style="width: 357px; height: 293px;" />
<pre>
<strong>Input:</strong> root = [3,4,5,1,3,null,1]
<strong>Output:</strong> 9
<strong>Explanation:</strong> Maximum amount of money the thief can rob = 4 + 5 = 9.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
</ul>
Tree; Depth-First Search; Dynamic Programming; Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function rob(root: TreeNode | null): number {
    const dfs = (root: TreeNode | null): [number, number] => {
        if (!root) {
            return [0, 0];
        }
        const [la, lb] = dfs(root.left);
        const [ra, rb] = dfs(root.right);
        return [root.val + lb + rb, Math.max(la, lb) + Math.max(ra, rb)];
    };
    return Math.max(...dfs(root));
}
