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LeetCode 198/213/337 House Robber 系列问题

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本文针对House Robber问题进行了探讨和研究，并给出了两种方法解决House Robber III问题

题目原文

337. House Robber III

Difficulty: Medium

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:
Input: [3,2,3,null,3,null,1]

1
2
3
4
5
  3
 / \
2   3
 \   \ 
  3   1

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

Example 2:
Input: [3,4,5,1,3,null,1]

1
2
3
4
5
    3
   / \
  4   5
 / \   \ 
1   3   1

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

解法

动态规划

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    public int rob(TreeNode root) {
        int[] result = helper(root);
        
        return Math.max(result[0], result[1]);
    }
    
    // 数组第1位表示使用root的值，第二位表示不用
    public int[] helper(TreeNode root) {
        if (root == null) {
            return new int[] {0, 0};
        }
        int[] left = helper(root.left);
        int[] right = helper(root.right);
        int[] result = new int[2];
        result[0] = root.val + left[1] + right[1];
        result[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        return result;
    }
}

记忆化搜索

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    Map<TreeNode, Integer> trueMap = new HashMap<>();
    Map<TreeNode, Integer> falseMap = new HashMap<>();
    public int rob(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return Math.max(helper(root, true), helper(root, false));
    }
    
    public int helper(TreeNode root, boolean grab) {
        if (root == null) {
            return 0;
        }
        if (grab) {
            if (trueMap.containsKey(root)) {
                return trueMap.get(root);
            } else {
                int val = root.val + helper(root.left, false) + helper(root.right, false);
                trueMap.put(root, val);
                return val;
            }
        } else {
            if (falseMap.containsKey(root)) {
                return falseMap.get(root);
            } else {
                int val = Math.max(helper(root.left, true), helper(root.left, false)) + 
                    Math.max(helper(root.right, true), helper(root.right, false));
                falseMap.put(root, val);
                return val;
            }
        }
    }
}