刷题337. House Robber III

一、题目说明

题目337. House Robber III，所有的房子连成二叉树状，不能“抢”直连的两个房间，请问最多可以抢到多少。难度是Medium！

二、我的解答

惭愧，这个题目思路始终不对。提交了n次也不正确，看了正确的思路：爷爷节点能偷到的最大钱=max(4个孙子偷的钱 + 爷爷的钱,两个儿子能偷的钱)

class Solution{
	public:
		//recursive + memo
		int rob(TreeNode* root){
			
			if(root==NULL){
				return 0;
			} 
			ump.clear();
			return	dfs(root);	
		}
		int dfs(TreeNode*root){
			if(root==NULL) return 0;
			if(ump.count(root)>0){
				return ump[root];
			}
			
			int money = root->val;
			if(root->left!=NULL){
				money += dfs(root->left->left) + dfs(root->left->right);
			}
			if(root->right!=NULL){
				money += dfs(root->right->left) + dfs(root->right->right);
			}
			int result = max(money,dfs(root->left)+dfs(root->right));
			ump[root] = result;
			return result;
		}
	private:
		unordered_map<TreeNode*,int> ump;
};

Runtime: 16 ms, faster than 81.99% of C++ online submissions for House Robber III.
Memory Usage: 23.1 MB, less than 55.56% of C++ online submissions for House Robber III.

三、优化措施

当然还有其他思路，略！

刷题337. House Robber III

原文：https://www.cnblogs.com/siweihz/p/12336507.html