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

  3
 / 
2   3
     
  3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

    3
   / 
  4   5
 /     
1   3   1

Maximum amount of money the thief can rob = 4 + 5 = 9.

这题一开始用递归 结果超时；随机加cache变成DP。。通过了

helper1表示抢root

helper2表示不抢root；此处注意不抢root时 即可以抢root.left和root.right 也可以不抢它们 二取其一 所以注意42-47行

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: The maximum amount of money you can rob tonight
     */
     Map<TreeNode, Integer> map1 = new HashMap<TreeNode, Integer>();
     Map<TreeNode, Integer> map2 = new HashMap<TreeNode, Integer>();
    public int houseRobber3(TreeNode root) {
        // write your code here
        if(root==null) return 0;
        return Math.max(helper1(root), helper2(root));
    }
    // include root
    private int helper1(TreeNode root){
        if(map1.containsKey(root)){
            return map1.get(root);
        }
        int sum = root.val;
        if(root.left!=null){
            sum += helper2(root.left);
        }
        if(root.right!=null){
            sum += helper2(root.right);
        }
        map1.put(root, sum);
        return sum;
    }
    // not include root
    private int helper2(TreeNode root){
        if(map2.containsKey(root)){
            return map2.get(root);
        }
        int sum = 0;
        if(root.left!=null){
            sum += Math.max(helper1(root.left), helper2(root.left));
        }
        if(root.right!=null){
            sum += Math.max(helper1(root.right), helper2(root.right));
        }
        map2.put(root, sum);
        return sum;
    }
}