LeetCode 337. House Robber III

原题链接在这里：https://leetcode.com/problems/house-robber-iii/

题目：

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.

题解：

List some examples and find out this has to be done with DFS.

One or null node is easy to think, thus use DFS bottom-up, devide and conquer.

Then it must return value on dfs. Each dfs needs current node and return [robRoot, notRobRoot], which denotes rob or skip current node.

robRoot = notRobLeft + notRobRight + root.val

notRobRoot = Math.max(robLeft, notRobLeft) + Math.max(robRight, notRobRight).

Note: when not robbint root, for left and right children. It could choose to rob child node or not.

Time Complexity: O(n).

Space: O(logn). 用了logn层stack.

AC Java:

/**
 * 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 = dfs(root);
        return Math.max(res[0], res[1]);
    }
    private int [] dfs(TreeNode root){
        int [] dp = new int[2];
        if(root == null){
            return dp;
        }
        int [] left = dfs(root.left);
        int [] right = dfs(root.right);
        //dp[0]表示偷root的, 那么左右都不能偷, 所以用left[1], right[1].
        dp[0] = left[1] + right[1] + root.val; 
        //dp[1]表示不偷root的, 那么左右偷不偷都可以, 取最大值
        dp[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        return dp;
    }
}

 类似House Robber, House Robber II.