337. House Robber III

    /*
     * 337. House Robber III
     * 2016-7-10 by Mingyang
     * I II都是线性的，都用的dp来做，这里我首先想到的也是dp，但是发现在Tree的结构里面表现的不是很合适
     * 后来我想到了前面做过的Largest BST Subtree，那道题目的话返回的是一个array，因为只能返回一样东西
     * 如果要返回两样的话，只能返回一个打包好的array，所以这道题目也是一样，最开始我想的是第一个返回的
     * 是1或者0，0表示我不取我本身，1表示我取我本身，第二个index返回的是这样做的偷的钱数
     * 是大神的思路，每个node都有两个状态，要么偷，要么不偷，第一个index表示如果root被偷了最多的钱
     * 第二个表示如果不被
     * Let's relate rob(root) to rob(root.left) and rob(root.right), etc. 
     * For the 1st element of rob(root), we only need to sum up the larger elements of rob(root.left) and rob(root.right), 
     * respectively, since root is not robbed and we are free to rob the left and right subtrees. 
     * For the 2nd element of rob(root), 
     * however, we only need to add up the 1st elements of rob(root.left) and rob(root.right),
     * respectively, plus the value robbed from "root" itself, 
     * since in this case it's guaranteed that we cannot rob the nodes of root.left and root.right.
     */
    public int rob(TreeNode root) {
        int[] res = robSub(root);
        return Math.max(res[0], res[1]);
    }
    private int[] robSub(TreeNode root) {
        if (root == null) {
            return new int[2];
        }
        int[] left = robSub(root.left);
        int[] right = robSub(root.right);
        int[] res = new int[2];
        res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        res[1] = root.val + left[0] + right[0];
        return res;
    }