[数据结构与算法]LeetCode系列337—打家劫舍 III

题意

337. 打家劫舍 III

题解

方法一：动态规划

class Solution {
public:
    unordered_map <TreeNode*, int> f, g;

    void dfs(TreeNode* node) {
        if (!node) {
            return;
        }
        dfs(node->left);
        dfs(node->right);
        f[node] = node->val + g[node->left] + g[node->right];
        g[node] = max(f[node->left], g[node->left]) + max(f[node->right], g[node->right]);
    }

    int rob(TreeNode* root) {
        dfs(root);
        return max(f[root], g[root]);
    }
};

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/house-robber-iii/solution/da-jia-jie-she-iii-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

struct SubtreeStatus {
    int selected;
    int notSelected;
};

class Solution {
public:
    SubtreeStatus dfs(TreeNode* node) {
        if (!node) {
            return {0, 0};
        }
        auto l = dfs(node->left);
        auto r = dfs(node->right);
        int selected = node->val + l.notSelected + r.notSelected;
        int notSelected = max(l.selected, l.notSelected) + max(r.selected, r.notSelected);
        return {selected, notSelected};
    }

    int rob(TreeNode* root) {
        auto rootStatus = dfs(root);
        return max(rootStatus.selected, rootStatus.notSelected);
    }
};

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/house-robber-iii/solution/da-jia-jie-she-iii-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。