# [Leetcode] 337. House Robber III | 打家劫舍 III

yPhantom 2020年02月05日 24次浏览

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:

``````Input: [3,2,3,null,3,null,1]

3
/ \
2   3
\   \
3   1

Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
``````

Example 2:

``````Input: [3,4,5,1,3,null,1]

3
/ \
4   5
/ \   \
1   3   1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
``````

Step by step tackling of the problem

``````public int rob(TreeNode root) {
int[] res = robSub(root);
return Math.max(res[0], res[1]);
}

private int[] robSub(TreeNode root) {
if (root == null) return new int[2];

int[] left = robSub(root.left);
int[] right = robSub(root.right);
int[] res = new int[2];

res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
res[1] = root.val + left[0] + right[0];

return res;
}
``````