[leetcode] House Robber III-优快云博客

本文链接：https://blog.youkuaiyun.com/sysubo/article/details/51232570

本文探讨了一种特殊情境下窃贼如何在不触动警报的情况下从二叉树结构的房屋中窃取最大金额的问题。通过定义二叉树节点并使用递归算法，实现了对每个节点的最优选择，确保相邻房屋不会同时被窃。最终给出两个示例，展示如何计算窃取的最大金额。

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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.

Example 1:

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

Example 2:

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

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
struct dpValue {
    dpValue(int v1 = 0, int v2 = 0) {
        takingVal = v1;
        notTakingVal = v2;
    }
    int takingVal;
    int notTakingVal;
};

class Solution {
public:
    int rob(TreeNode* root) {
        dpValue ans = solve( root );
        return (ans.takingVal > ans.notTakingVal)? ans.takingVal : ans.notTakingVal;
    }
    
    dpValue solve( TreeNode *root ) {
        if( root == NULL )
            return dpValue();
        dpValue cur( root->val, 0 );
        dpValue leftVal = solve(root->left);
        dpValue rightVal = solve(root->right);
        cur.takingVal += (leftVal.notTakingVal+rightVal.notTakingVal);
        cur.notTakingVal += ( (leftVal.takingVal > leftVal.notTakingVal)? leftVal.takingVal : leftVal.notTakingVal );
        cur.notTakingVal += ( (rightVal.takingVal > rightVal.notTakingVal)? rightVal.takingVal : rightVal.notTakingVal );
        return cur;
    }

};