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.
此题是house robber的变体，解法与之前的解法类似，也是动态规划方法解决（每个子问题都和原问题相同，只是规模不同，并且满足最优子结构和重叠子问题条件）。不同之处在于，这里是树，而之前给的是数组，这里面需要用到深度优先遍历，用res数组来记录到该节点时候，最大值是多少，res[0]表示不包括该节点，而res[1]表示包括该节点，代码如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int rob(TreeNode root) {
int[] res = subrob(root);
return Math.max(res[0],res[1]);
}
public int[] subrob(TreeNode root){
if(root==null) return new int[2];
int[] left = subrob(root.left);
int[] right = subrob(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;
}
}