LeetCode 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.

 

思路

rob兩種狀況：1.取當前節點以及當前節點的孫子節點的值的和 2.取當前節點的孩子節點的值的和。而後取二者最大值.

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
        val = x;
    }
}
public class Solution {

    public int helprob(Map<TreeNode, Integer> map, TreeNode root) {
        
        if(root == null){
            return 0;
        }
        
        if(map.containsKey(root)){
            return map.get(root);
        }
        int res = 0;
        if(root.left != null){
            res += helprob(map, root.left.left) + helprob(map, root.left.right);
        }
        
        if(root.right != null){
            res += helprob(map, root.right.left) + helprob(map, root.right.right);
        }
        
        res = Math.max(root.val + res, helprob(map, root.left) + helprob(map, root.right));
        map.put(root, res);
        return res;
    }
    
    
    public int rob(TreeNode root){
        //定義一個map來存儲取每一個節點的最大值
        Map<TreeNode, Integer> map = new HashMap<>();
        return helprob(map, root);
    }

}