题目描述:
![](https://img.laitimes.com/img/__Qf2AjLwojIjJCLyojI0JCLiAnYldHL0FWby9mZvwFN4ETMfdHLkVGepZ2XtxSZ6l2clJ3LcV2Zh1Wa9M3clN2byBXLzN3btgHL9s2RkBnVHFmb1clWvB3MaVnRtp1XlBXe0xCMy81dvRWYoNHLwEzX5xCMx8FesU2cfdGLwMzX0xiRGZkRGZ0Xy9GbvNGLpZTY1EmMZVDUSFTU4VFRR9Fd4VGdsQTMfVmepNHLrJXYtJXZ0F2dvwVZnFWbp1zczV2YvJHctM3cv1Ce-cmbw5SNyYDO1MGOxUWYyUWY4AjNzYzX0MjN1ATMxAzLcFTMyIDMy8CXn9Gbi9CXzV2Zh1WavwVbvNmLvR3YxUjLyM3Lc9CX6MHc0RHaiojIsJye.png)
思路1:暴力求解。当前节点能偷到的最大钱数有两种可能的计算方式。1.爷爷节点和孙子节点偷的钱的总和 2.儿子节点偷的钱的总和。则当前节点能偷到的最大钱数只要取一个max即可。
/**
* 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:
int rob(TreeNode* root) {
if(root==NULL)
return 0;
int money=root->val;
if(root->left){
money+=rob(root->left->left)+rob(root->left->right);
}
if(root->right){
money+=rob(root->right->left)+rob(root->right->right);
}
return max(money,rob(root->left)+rob(root->right));
}
};
时间复杂度太高,超时。
思路2:带有备忘录的遍历
/**
* 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:
int rob(TreeNode* root) {
if(root==NULL)
return 0;
if(mp.find(root)!=mp.end())
return mp[root];
int money=root->val;
if(root->left){
money+=rob(root->left->left)+rob(root->left->right);
}
if(root->right){
money+=rob(root->right->left)+rob(root->right->right);
}
mp.insert(make_pair(root,max(money,rob(root->left)+rob(root->right))));
return mp[root];
}
private:
map<TreeNode*,int> mp;
};
/**
* 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:
int rob(TreeNode* root) {
if(root==NULL)
return 0;
vector<int> steal(2,0);
helper(root,steal);
return max(steal[0],steal[1]);
}
void helper(TreeNode* root,vector<int>& steal){
if(root==NULL)
return ;
vector<int> left(2,0);
vector<int> right(2,0);
helper(root->left,left);
helper(root->right,right);
steal[0]=max(left[0],left[1])+max(right[0],right[1]);
steal[1]=left[0]+right[0]+root->val;
}
};