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:
\2 3
\ \
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
4 5
/ \ \
Maximum amount of money the thief can rob = 4 + 5 = 9.
思路：
题意：不能连续抢直接相连的两个节点。即例2中，抢了3就不能抢4，5。问最多能抢好多。
给一个二叉树，求它不直接相连的节点的val和最大为多少。
如果抢了当前节点，那么它的左右孩子就肯定不能抢了。
如果没有抢当前节点，左右孩子抢不抢取决于左右孩子的孩子的val大小。
像House Robber I一样，使用动态规划法，对于每个节点，使用两个变量，res[0], res[1]，分别表示不选择当前节点子树的数值和，选择当前节点子树的数值和，动态规划的思想，然后递归。
/*** 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 = robSub(root);
return Math.max(res[0], res[1]);
}
public 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]); // do not choose current node
res[1] = root.val + left[0] + right[0]; // choose current node
return