[Leetcode] 337. House Robber III 解题报告

题目：

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：

二叉树一般情况下用递归是最好的办法。分析本题目可知，要想获得最大值有两种选择：

1）抢劫根节点，那么此时就不能抢劫左右子树的根节点；

2）不抢劫根节点，那么此时就可以抢劫左右子树的根节点。

两种情况中取受益最大者即可。在实现中值得注意的是，在递归过程中有可能会引起重复计算。例如代码片段1中，我们在rob(root->left)的调用过程中会调用rob(root->left->left)，而这个调用在rob(root)中也出现，导致大量重复计算。一种比较好的递归方式见代码片段2，我们采用“首递归”，并用传递引用的方式记录左右子树已经取得的最大值，可以完美地避免重复计算。（测试结果表明代码片段2的效率比代码片段1的效率高100倍左右）

代码：

1、低效递归：

/** * 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) {            return 0;        }        int rob_root = root->val;        if (root->left) {            rob_root += rob(root->left->left);            rob_root += rob(root->left->right);        }        if (root->right) {            rob_root += rob(root->right->left);            rob_root += rob(root->right->right);        }        int not_rob_root = rob(root->left);        not_rob_root += rob(root->right);        return max(rob_root, not_rob_root);    }};

2、高效递归：

/** * 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 left = 0, right = 0;        return rob(root, left, right);    }private:    int rob(TreeNode* root, int &left, int &right) {        if(!root) {            return 0;        }        int left_left = 0, left_right = 0, right_left = 0, right_right = 0;        left = rob(root->left, left_left, left_right);      // left_left and left_right will be updated during recursions        right = rob(root->right, right_left, right_right);  // right_left and right_right will be updated during recursions            return max(root->val + left_left + left_right + right_left + right_right, left + right);    }};