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:

Input: [3,2,3,null,3,null,1]

     3
    / \
   2   3
    \   \ 
     3   1

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

Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
   4   5
  / \   \ 
 1   3   1

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

又是个house rober的题目，前几天基本都是使用动态规划来做的，这题变成了二叉树，而且不能偷相邻节点。

这题没想到好方法，直接暴力解题。大概思路就是保存每个节点能偷到的最大数量，当然这个数量可能是包含该节点的，或者是不包含该节点的。

当前节点的最大数量，为当前节点加上相邻的左右节点的子节点的最大值，不包含当前节点的相邻左右节点的最大值相比较，取最大值：

    int rob(TreeNode* root) {
        unordered_map<TreeNode*, int> um;
        return dfs(root, um);
    }

    static int dfs(TreeNode* root, unordered_map<TreeNode*, int> &rvm) {
        if (nullptr == root) {
            return 0;
        }
        if (rvm.count(root) != 0) {
            return rvm[root];
        }
  
        int nVal = root->val;
        if (nullptr != root->left) {
            nVal += dfs(root->left->left, rvm);
            nVal += dfs(root->left->right, rvm);
        }
        if (nullptr != root->right) {
            nVal += dfs(root->right->left, rvm);
            nVal += dfs(root->right->right, rvm);
        }
        nVal = std::max(nVal, dfs(root->left, rvm) + dfs(root->right, rvm));

        rvm[root] = nVal;
        return nVal;
    }