[LeetCode] 337. House Robber III

Problem:

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.

Solution:

/** * 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) {        int l = 0, r = 0;        int max = dfsRob(root, l, r);        return max;    }        /* 对每颗子树：      计算其根节点 + 四个孙子节点的值（若不存在即为0） root->val + ll + lr + rl + rr      计算其左子节点值 + 右子节点的值 l + r      比较并返回较大者*/    int dfsRob(TreeNode* root, int& l, int& r) {        if (root == NULL) return 0;        int ll = 0, lr = 0, rl = 0, rr = 0;        l = dfsRob(root->left, ll, lr);        r = dfsRob(root->right, rl, rr);        return max(root->val + ll + lr + rl + rr, l + r);    }};

关键点说明：对每颗子树来说，有两个各不相邻的节点的集合（假设节点均不为空）：

           {根节点，四个孙子节点} 和 {左子节点， 右子节点} 

所以，只需要比较这两个集合的节点值的和即可得到每颗子树的要求的最大值，通过递归可以得到最终的最大值。