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.
例子
例子 1
Input:
[3,2,3,null,3,null,1]Output: 7 Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
例子 2
Input:
[3,4,5,1,3,null,1]Output: 9 Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
解题思路
这道题可以用动态规划的思路来做，对于一个根结点 root 而言，情况有两种：
- 使用相隔一层之后的结果加上当前节点的值
- 直接使用相邻层的结果
因此转移方程为：
rob(root) = max(root->val + rob(two childrens of (root->left)) + rob(two children of root->right), rob(root->left) + rob(root->right))
为了避免重复计算，我们可以使用一个哈希表来存储每个节点的 rob 结果。
代码如下：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
#include <unordered_map>
class Solution {
public:
int rob(TreeNode* root) {
if (!root) return 0;
if (hmap.find(root) != hmap.end()) return hmap[root];
int curr_sum = root->val;
if (root->left) {
curr_sum += rob(root->left->left);
curr_sum += rob(root->left->right);
}
if (root->right) {
curr_sum += rob(root->right->left);
curr_sum += rob(root->right->right);
}
int prev_sum = rob(root->left) + rob(root->right);
hmap[root] = max(curr_sum, prev_sum);
return hmap[root];
}
private:
std::unordered_map<TreeNode*, int> hmap;
};
- 时间复杂度: O(n)
- 空间复杂度: O(n)
GitHub 代码同步地址： 337.HouseRobberIII.cpp
其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions