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.
题目含义：不能连续抢直接相连的两个节点。即例2中，抢了3就不能抢4，5。问最多能抢好多。
思路：给一个二叉树，求它不直接相连的节点的val和最大为多少。
如果抢了当前节点，那么它的左右孩子就肯定不能抢了。
如果没有抢当前节点，左右孩子抢不抢取决于左右孩子的孩子的val大小。
1 private int[] dfs(TreeNode root) 2 { 3 int[] result = {0,0}; //result[0]表示抢当前节点 result[1]表示不抢当前节点 4 if (root == null) return result; 5 int[] leftResult = dfs(root.left); 6 int[] rightResult = dfs(root.right); 7 result[0] = leftResult[1] +rightResult[1] + root.val;//抢了当前节点，它的左右孩子就不可以抢了 8 result[1] = Math.max(leftResult[0],leftResult[1]) + Math.max(rightResult[0],rightResult[1]);//不抢当前节点，左右孩子可抢可不抢 9 return result; 10 } 11 12 public int rob(TreeNode root) { 13 int[] result = dfs(root); 14 return Math.max(result[0],result[1]); 15 }
