337 House Robber III

337. House Robber III

1. Question

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:
3
/ \
2 3
\ \
3 1
Maximum amount of money the thief can rob =
3 + 3 + 1 = 7.
Example 2:
3
/ \
4 5
/ \ \
1 3 1
Maximum amount of money the thief can rob =
4 + 5 = 9.

2. Implementation

(1) DFS
public class Solution {
public int rob(TreeNode root) {
Map<TreeNode, Integer> map = new HashMap<>();
return robHouse(root, map);
}
public int robHouse(TreeNode node, Map<TreeNode, Integer> map) {
if (node == null) {
return 0;
}
if (map.containsKey(node)) {
return map.get(node);
}
int val = 0;
if (node.left != null) {
val += robHouse(node.left.left, map) + robHouse(node.left.right, map);
}
if (node.right != null) {
val += robHouse(node.right.left, map) + robHouse(node.right.right, map);
}
int res = Math.max(val + node.val, robHouse(node.left, map) + robHouse(node.right, map));
map.put(node, res);
return res;
}
}

3. Time & Space Complexity

DFS: 时间复杂度O(n), 空间复杂度O(n),因为用了map