[337]House Robber III

最新推荐文章于 2019-09-24 09:26:39 发布

原创最新推荐文章于 2019-09-24 09:26:39 发布 · 275 阅读

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本文探讨了一道经典的二叉树遍历问题——如何在不触动报警机制的情况下，计算出夜间抢劫的最大收益。通过递归策略，我们为每个节点定义了一个子函数，该函数返回两个值：抢当前房屋的最大收益和跳过当前房屋的最大收益。

【题目描述】

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:

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

Example 2:

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

【解题思路】

发现自己在搜索这种题型上还是欠缺的，参考了discuss里一个很详细的解释：

https://discuss.leetcode.com/topic/39834/step-by-step-tackling-of-the-problem

【代码】

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> robsub(TreeNode* root){
         if (root == NULL) {
            return vector<int>(2,0);
        }
        vector<int> left = robsub(root->left);
        vector<int> right = robsub(root->right);
        vector<int> ans(2,0);
        ans[0] = max(left[0],left[1]) + max(right[0], right[1]);
        ans[1] = root->val + left[0] + right[0];
        return ans;
    }
    int rob(TreeNode* root) {
       vector<int> money = robsub(root);
       return max(money[0],money[1]);
    }
    
};