Java [Leetcode 337]House Robber III

题目描述：

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.

解题思路：

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。

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 = 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 res;
    }
}