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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

树形dp 

• rob_root = max(rob_L + rob_R , no_rob_L + no_nob_R + root.val)
• no_rob_root = rob_L + rob_R

• /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
   //tree DP
  public class Solution {
      public int rob(TreeNode root) {
       return dfs(root)[0];
      }
      public int[] dfs(TreeNode root){
          if(root == null) return new int[2];
          int[] left = dfs(root.left);
          int[] right = dfs(root.right);
          int dp[] = new int[2];
          dp[1] = left[0] + right[0];
          dp[0] = Math.max(dp[1] , left[1] + right[1] + root.val);
          return dp;
      }
  }