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.

递归

一看到树的问题，首先就想到了divide and conque和递归。

对每一个节点，都存在两种状态，一种是偷，一种是不偷。 父节点可以由子节点推出，

  1. 当要偷父节点的时候，子节点一定不能偷
  2. 当不偷父节点的时候，子节点可以偷也可以不偷
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class ReturnType {
    int stealRoot = 0;
    int notStealRoot = 0;
    public ReturnType(int stealRoot, int notStealRoot) {
        this.stealRoot = stealRoot;
        this.notStealRoot = notStealRoot;
    }
}
public int rob(TreeNode root) {
    ReturnType ans = helper(root);
    return Math.max(ans.stealRoot, ans.notStealRoot);
}
public ReturnType helper(TreeNode root) {
    if (root == null) {
        return new ReturnType(0, 0);
    }
    ReturnType left = helper(root.left);
    ReturnType right = helper(root.right);
    return new ReturnType(left.notStealRoot + right.notStealRoot + root.val, Math.max(left.stealRoot, left.notStealRoot) + Math.max(right.stealRoot, right.notStealRoot));
}