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Monday, July 24, 2017

[Leetcode] House Robber III, Solution

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.


[Thoughts]

这就可以归结为树的递归遍历问题。对于任意一个节点,如下图三角虚线内,该子树的最大值一定是下面两个可能:

  1. root + left_left + left_right + right_left + right_right
  2. left + right
所以对于任意一个节点,只要递归的调用这个关系,就可以得到解。





[Code]

1:    int rob(TreeNode* root) {  
2:      if(root == NULL) return 0;  
3:        
4:      int val = 0;  
5:        
6:      if(root->left) {  
7:        val += rob(root->left->left) + rob(root->left->right);  
8:      }  
9:        
10:      if(root->right) {  
11:        val += rob(root->right->left) + rob(root->right->right);  
12:      }  
13:    
14:      return max(root->val + val, rob(root->left) + rob(root->right));   
15:    }  


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