具体思想:
自己第一次想的是自顶向下进行判断,该节点选或者不选,进行爆搜;
但是发现爆时间;
后来发现存在递归子问题,从底向上搜即可,同时使用map记录;
题解写的更简单;
具体代码:
1.My version:
/**
* 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) {}
* };
*/
class Solution {
public:
int rob(TreeNode* root) {
preorder(root);
return record[root];
}
void preorder(TreeNode* root){
if(!root)
return;
preorder(root->left);
preorder(root->right);
if(!root->left&&!root->right){
//如果是叶子节点;
record[root]=root->val;
return;
}
//如果不选择该节点;
int count=0;
count+=root->left?record[root->left]:0;
count+=root->right?record[root->right]:0;
//如果选择该节点;
int temp=cnt(root->left);
temp+=cnt(root->right);
count=max(count,temp+root->val);
record[root]=count;
return;
}
int cnt(TreeNode* root){
if(!root)
return 0;
int count=0;
count+=root->left?record[root->left]:0;
count+=root->right?record[root->right]:0;
return count;
}
int dfs(TreeNode* root,bool flag){
if(!root){
return 0;
}
if(!flag){
//当前节点不能选择;
return dfs(root->left,true)+dfs(root->right,true);
}else{
//当前节点可以选择;
//选了,或者可以选但是不选;
int ret=0;
ret=max(ret,dfs(root->left,false)+dfs(root->right,false)+root->val);//选择当前节点;
ret=max(ret,dfs(root->left,true)+dfs(root->right,true));//不选择当前节点;
return ret;
}
}
unordered_map<TreeNode*,int>record;//记录子树信息;
};
2.官方 version:
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]);
}
};