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;
     }
 }