leetcode解题: House Robber III (337)

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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:

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5
  3
 / \
2   3
 \   \
  3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:

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2
3
4
5
    3
   / \
  4   5
 / \   \
1   3   1

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

解法1：

参考了这个帖子的解法3。 对于每一个节点，存在两种可能， 取或者不取。 那么可以用一个帮助函数，每一次计算的时候保存这两种情况的值。
用一个vector保存，v[0]表示从这个点出发，不包含这个点最大的可取值， v1表示从这个点出发并且包含这个点可取的最大值。
那么对于一个root， 分别计算left和right的vector，再按取或者不取的情况计算出root的vector。
如果取root的值，那一定不能取left和right的值， 所以最大取值是left1 + right1 + val
如果不取root的值，那可以取left和right的值， 这也等价于两颗树，对这两颗树取这个问题的最大值，就是left两种情况的最大值+right两种情况的最大值。
C++

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/**
 * 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) {
        vector<int> temp = helper(root);
        return max(temp[0], temp[1]);
    }
    vector<int> helper(TreeNode* root) {
        if (!root) {
            return vector<int>(2, 0);
        }
        vector<int> left = helper(root->left);
        vector<int> right = helper(root->right);
        vector<int> res(2,0);
        res[0] = max(left[0], left[1]) + max(right[0], right[1]);
        res[1] = left[0] + right[0] + root->val;
        return res;
    }
};

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    class MarkedTreeNode {
        int take;
        int notake;
        public MarkedTreeNode(int take, int notake) {
            this.take = take;
            this.notake = notake;
        }
    };
    public int rob(TreeNode root) {
        if (root == null) return 0;
        MarkedTreeNode res = helper(root);
        return Math.max(res.take, res.notake);
    }
    private MarkedTreeNode helper(TreeNode root) {
        if (root == null) return new MarkedTreeNode(0,0);
        MarkedTreeNode left = helper(root.left);
        MarkedTreeNode right = helper(root.right);
        int take = root.val + left.notake + right.notake;
        int notake = Math.max(left.notake, left.take) + Math.max(right.notake, right.take);
        return new MarkedTreeNode(take, notake);
    }
}