leetcode——337—— House Robber III

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

Example 1:

     3
    / \
   2   3
    \   \ 
     3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

     3
    / \
   4   5
  / \   \ 
 1   3   1

Maximum amount of money the thief can rob = 4 + 5 = 9.

递归函数返回一个大小为2的一维数组res，其中res[0]表示抢劫当前节点值的最大值，res[1]表示不抢劫当前节点的最大值，那么我们在遍历某个节点时，首先对其左右子节点调用递归函数，分别得到包含与不包含左子节点值的最大值，和包含于不包含右子节点值的最大值，那么当前节点的res[1]就是左子节点两种情况的较大值加上右子节点两种情况的较大值，res[0]就是不包含左子节点值的最大值加上不包含右子节点值的最大值，和当前节点值之和，返回即可

/**
 * 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) {
       vector<int> res=dfs(root);
       return max(res[0],res[1]);
        
        
    }
    vector<int> dfs(TreeNode *root){
        vector<int> res(2,0);//0 存储robed 最大利益,  1,nonrobed 最大利益
        if(!root)
            return res;
        
        vector<int> lres=dfs(root->left);
        vector<int> rres=dfs(root->right);
        
        res[0]=lres[1]+rres[1]+root->val;
        
        res[1]=max(lres[0],lres[1])+max(rres[0],rres[1]);
        
        return res;
    }
    
};