/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    //对于每一个节点，都有选或者不选两种决定
    //选了该节点，则它的左子节点和右子节点不能选
    //设定一个数组，下标0代表选了该节点的最大偷盗金额，下标1代表未选择该节点的最小偷盗金额
    // dfs + dp
    // 时间复杂度O(n),空间复杂度O(n); 每个节点遍历一次
    public int rob(TreeNode root) {
        int[] res = dfs(root); // res[]记录选root和不选root可以偷窃的最大值
        return Math.max(res[0], res[1]);
    }
    public int[] dfs(TreeNode root) {
        if (root == null) {
            return new int[]{0,0};
        }
        int[] left = dfs(root.left);
        int[] right = dfs(root.right);
        int selected = root.val + left[1] + right[1];
        int noSelected = Math.max(left[0], left[1]) + Math.max (right[0], right[1]);
        return new int[]{selected, noSelected};
    }
}