337. 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:

     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;

    }

}