[LeetCode] 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.

化成递归问题，如果抢劫root，那么结果位 root->val + rob(root.child.child)

如果不抢劫，那么结果为rob(root.child). 两者中取最大值。

特别注意写递归时要避免重复递归的情形，不然会TLE。

采用备忘录方法，对于这个题，有更简单的写法。

/** * 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;        int l=0;int r=0;        return recursion(root,l,r);    }    int recursion(TreeNode* root,int &l,int &r) //记录以左右子树节点做根的子树大小    {        if(!root)            return 0;        int ll=0; int lr = 0;int rl = 0;int rr = 0; //记录子节点        l = recursion(root->left,ll,lr);        r = recursion(root->right,rl,rr);        return max(root->val+ll+lr+rl+rr,l+r);    }    };