[LintCode/LeetCode] House Robber III

macg0406 发布于2019-08-14 15:34 / 1125人阅读

摘要：解法真的非常巧妙，不过这道题里仍要注意两个细节。中，为时，返回长度为的空数组建立结果数组时，是包括根节点的情况，是不包含根节点的情况。而非按左右子树来进行划分的。

Problem

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

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

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

Note

DFS解法真的非常巧妙，不过这道题里仍要注意两个细节。
DFS中，root为null时，返回长度为2的空数组；
建立结果数组A时，A[0]是包括根节点的情况，A[1]是不包含根节点的情况。而非按左右子树来进行划分的。

Solution

public class Solution {
    public int houseRobber3(TreeNode root) {
        int[] A = dfs(root);
        return Math.max(A[0], A[1]);
    }
    public int[] dfs(TreeNode root) {
        if (root == null) return new int[2];
        int[] left = dfs(root.left);
        int[] right = dfs(root.right);
        int[] A = new int[2];
        A[0] = left[1] + root.val + right[1];
        A[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        return A;
    }
}

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