337 House Robber III Java Solutions

The thief has found himself a new place for his thievery again. There is a entrance to this area, called the "root." Besides the root, each house have one and only one parent house. After a tour, the Smart thief realized the ' all houses in this ' place forms a binary tree. It would automatically contact the police if and 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.

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1 /**2 * Definition for a binary tree node.3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One      Public intRob (TreeNode root) { A         int[] res =Robdfs (root); -         returnMath.max (res[0],res[1]); -     } the      -      Public int[] Robdfs (TreeNode root) { -         if(Root = =NULL)return New int[2]; -         int[] left =Robdfs (root.left); +         int[] right =Robdfs (root.right); -         int[] ans =New int[2]; +Ans[0] = left[1] + root.val + right[1]; AANS[1] = Math.max (left[0], left[1]) + Math.max (right[0],right[1]); at         returnans; -     } -}

In Dfs, when Root is null, an empty array of length 2 is returned;
When the resulting array res is established, Res[0] is the case that includes the root node, and res[1] is not the root node. Instead of dividing it by the left and right sub-trees.

337 House Robber III Java Solutions