# House Robber III

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.
``````

## Analysis

the two conditions for dynamic programming: "optimal substructure" + "overlapping of subproblems", we actually have a DP problem

## Solution

DP with Memorization O(n) space cost (n is the total number of nodes; stack cost for recursion is not counted).

``````public int rob(TreeNode root) {
return robSub(root, new HashMap<>());
}

private int robSub(TreeNode root, Map<TreeNode, Integer> map) {
if (root == null) return 0;
if (map.containsKey(root)) return map.get(root);

int val = 0;

if (root.left != null) {
val += robSub(root.left.left, map) + robSub(root.left.right, map);
}

if (root.right != null) {
val += robSub(root.right.left, map) + robSub(root.right.right, map);
}

val = Math.max(val + root.val, robSub(root.left, map) + robSub(root.right, map));
map.put(root, val);

return val;
}
``````

DP (0ms 100% AC)

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int rob(TreeNode root) {
int[] res = robHelper(root);
return Math.max(res[0], res[1]);
}
int[] robHelper(TreeNode root) {
if (root == null) return new int[2];
int[] left = robHelper(root.left);
int[] right = robHelper(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;
}
}
``````