leetcode337——打家劫舍III

Posted 2022-12-11 dtwd886

方法一：

树上dp,通过两个unordered_map记录每个点选择获取或者不选择获取的状态。

选择则状态转移方程为f[o]=o->val+g[o->left]+g[o->right]

不选择点的状态为g[o]=max(f[o->left],g[o->left])+max(f[o->right],g[o->right])

/**
 * 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:
    unordered_map<TreeNode*,int>f,g;//f代表选择当前结点，o代表不选择当前节点
    void DFS(TreeNode* o)
    
        if(!o)return;
        DFS(o->left);
        DFS(o->right);
        f[o]=o->val+g[o->left]+g[o->right];
        g[o]=max(f[o->left],g[o->left])+max(f[o->right],g[o->right]);
    
    int rob(TreeNode* root) 
        DFS(root);
        return max(f[root],g[root]);
    
;

方法二：

由于父结点的状态只和子节点有关，所以产生可以省去hash表的方法。

/**
 * 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:
    struct SelectItem
    
        int selected;
        int notSelected;
    ;
    SelectItem DFS(TreeNode* o)
    
        if(!o)return 0,0;
        auto l=DFS(o->left);
        auto r=DFS(o->right);
        int select=o->val+l.notSelected+r.notSelected;
        int notSelect=max(l.selected,l.notSelected)+max(r.selected,r.notSelected);
        return select,notSelect;
    
    int rob(TreeNode* root) 
        auto result=DFS(root);
        return max(result.selected,result.notSelected);
    
;