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.

一个二叉树，不能取相邻的两个数，求可以取到的树之和的最大值

rob[2]中 rob[0]代表rob节点所能得到的最大值，rob[1]代表从rob的子节点到叶子节点所能得到的最大值。用类似深度优先的遍历，动态规划得出结果

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int Rob(TreeNode root) {
        return dfs(root)[0];
    }

    public int[] dfs(TreeNode root)
    {
        int[] rob = {0, 0};
        if(root != null)
        {
            int[] leftRob = dfs(root.left);
            int[] rightRob = dfs(root.right);
            rob[1] = leftRob[0] + rightRob[0];
            rob[0] =  System.Math.Max(rob[1], leftRob[1] + rightRob[1] + root.val);
        }
        return rob;
    }
}