# 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, res);
}
int[] robHelper(TreeNode root) {
if (root == null) return new int;
int[] left = robHelper(root.left);
int[] right = robHelper(root.right);
int[] res = new int;
res = Math.max(left, left) + Math.max(right, right);
res = root.val + left + right;
return res;
}
}