class Solution {
public:int rob(vector<int>& nums) {vector<int> dp(nums.size()); //考虑下标i的最大偷窃金额dp[i]if(nums.size()==0) return 0;if(nums.size()==1) return nums[0];dp[0] = nums[0];dp[1] = max(nums[0],nums[1]);for(int i = 2; i<nums.size();i++){dp[i] = max(dp[i-1],dp[i-2]+nums[i]); //不偷i:dp[i-1],偷i:dp[i-2]+nums[i]}return dp[nums.size()-1];}
};

class Solution {
public:int lineResult(vector<int>& nums,int start,int end){ //start和end都是可以取到的Index，两者数据量start-end+1if(start == end) return nums[start]; //此时数据量为1vector<int> dp(nums.size());dp[start] = nums[start];dp[start+1] = max(nums[start],nums[start+1]);for(int i = start+2; i<= end;i++){ //可以取到end故i<= enddp[i] = max(dp[i-2]+nums[i],dp[i-1]);}return dp[end];}int rob(vector<int>& nums) {if (nums.size() == 0) return 0;if (nums.size() == 1) return nums[0]; //这两行确保了nums.size()最小为2int result1 = lineResult(nums,0,nums.size()-2);int result2 = lineResult(nums,1,nums.size()-1);return max(result1,result2);}
};

class Solution {
public:vector<int> robTree(TreeNode* cur){if(cur == nullptr) return vector<int> {0,0}; //dp[0]为不偷，dp[1]为偷vector<int> leftdp = robTree(cur->left); //左右中后续遍历vector<int> rightdp = robTree(cur->right);int val1 = max(leftdp[0],leftdp[1]) + max(rightdp[0],rightdp[1]); //不偷该处节点，那么取左节点和右节点最大值int val2 = cur->val + leftdp[0] + rightdp[0];return {val1,val2};}int rob(TreeNode* root) {vector<int> dproot = robTree(root);return max(dproot[0],dproot[1]);}
};