337. 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.

题意：如之前的house robberI & II，假设房子的构造是二叉树型的，父子结点不能同时偷窃，则问最大的偷窃方案。

思路：

1. 动态规划的思想是通过（存储）局部的最优解，在局部最优解的基础上，通过进一步决策，达到全局的最优解。当然，局部面临的问题与全局面临的问题具有相似性。

2. 以该题为例，求解这棵树root的最优解，那么我假设左子树和右子树的最优解情况我已经知道，那么我想知道整颗树的最优解，就需要知道取root的val时，最优解是多少，还要知道不取root->val时的最优解多少，从而比较即可知道整颗树的最优解。取root->val时，最优解=不取左子树跟结点时，左子树的最优解+不取右子树的跟结点时右子树的最优解+root->val；不取root->val时，最优解=左子树的最优解+右子树的最优解。左子树的最优解=max{ 取左子树根结点时最优解， 不取左子树跟结点时最优解 }。

3. 同理局部子问题（以左子树为例），左子树也是一颗树，也存在左子树的根结点取和不取时最优的问题，假设左子树跟结点取和不取的最优值分别为left, nleft；右子树的对应值right，nright

4. 那么整棵树的取和不取时的最优值分别别表示为：with=nleft+nright+root->val, without=max(left, nleft)+max(right, nright)

5. 决策的时候即是比较大小即可。子问题同样如此递归处理。

/**
 * 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 with, without;
		dfs(root, with, without);
		return max(with, without);
	}

	void dfs(TreeNode* root, int& with, int& without){
		if (root == NULL){
			with = 0;
			without = 0;
			return;
		}

		int left, nleft;
		int right, nright;

		dfs(root->left, left, nleft);
		dfs(root->right, right, nright);

		with = nleft + nright + root->val;
		without = max(left, nleft) + max(right, nright);
	}	
};