## House Robber III

07/18/2016 Tree Depth First Search

## Question

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
/ \
2   3
\   \
3   1


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

## Solution

Result: Accepted Time: 15 ms

Here should be some explanations.

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

struct State{
int a,b;
int get_max()
{
return a > b ? a : b;
}
State():a(0),b(0){}
State(int aa,int bb):a(aa),b(bb){}
};

State robber(TreeNode * root)
{
if(root == NULL)
return State();
State left = robber(root->left);
State right = robber(root->right);
return State(left.get_max() + right.get_max(), left.a + right.a + root->val);
}
class Solution {
public:
int rob(TreeNode* root) {
return robber(root).get_max();
}
};


Complexity Analytics

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$