Leetcode-337 House Robber III

337. House Robber III

Medium

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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.

/**
 * 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) {
        if(root == NULL){
            return 0;
        }
        if(root->left == NULL && root->right == NULL){
            return root->val;
        }
        int l = rob(root->left);
        int r = rob(root->right);
        int ll,lr,rl,rr;
        if(root->left && root->left->left){
            ll = rob(root->left->left);
        }
        else{
            ll = 0;
        }
        if(root->left && root->left->right){
            lr = rob(root->left->right);
        }
        else{
            lr = 0;
        }
        if(root->right && root->right->left){
            rl = rob(root->right->left);
        }
        else{
            rl = 0;
        }
        if(root->right && root->right->right){
            rr = rob(root->right->right);
        }
        else{
            rr = 0;
        }
        
        return max(l+r, root->val+ll+lr+rr+rl);
        
        
    }
    
    int max(int a, int b){
        if(a>b){
            return a;
        }
        else{
            return b;
        }
    }
};

 

一直在理解错题意，首先是觉得答案应该是偶数行的value或者奇数行的value总和。

后来甚至开始写背包问题，每行的value和为背包中的元素，算最大值

但是忘记考虑了，相邻行但不是父子关系的也可以

其实就应该用递归来解决问题，在highlevel上思考就好了。