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.

【思路】拿root(第0层)为例,如果取第0层的节点,则第1层的节点不能取;如果不取第0层的节点,则第1层的节点可取可不取。这样我们需要记录下每个节点取与不取时能够获取的最大钱数,通过深度优先遍历二叉树,最后取root节点返回的两个数值中大的就可以了。

<pre name="code" class="cpp">/**
 * 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:
    vector<int> robSub(TreeNode* root){
        if(root==NULL) return vector<int>(2,0);;
        
        vector<int> left = robSub(root->left);
        vector<int> right = robSub(root->right);
        
        vector<int> re(2,0);
        re[0] = max(left[0],left[1]) + max(right[0],right[1]);
        re[1] = root->val + left[0] + right[0];
        
        return re;
    }
    int rob(TreeNode* root) {
        vector<int> re = robSub(root);
        return max(re[0], re[1]);
    }
};


 
 

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