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