LeetCode OJ 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.

对于这个题目我们有什么思路呢？我们从根节点开始有两种选择，要么偷取根节点的值，要么不偷取根节点。偷取根节点的话，我们接下来只能偷取根节点孩子节点的孩子节点。如果不偷取根节点的话，我们可以直接偷取根节点的孩子节点。然后我们把这两种方式得到的值进行比较，取较大的那个。因此这是一个递归的过程：

rob(root) = max{rob(root.left) + rob(root.rght), root.val + rob(root.left.left) + rob(root.left.right) + rob(root.right.left) + rob(root.right.right)}

if(root == null) rob(root) = 0;

if(root.left == null && root.right == null) rob(root) = root.val

有了上面的递归表达式，我们就很容易进行编程啦！代码如下：

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public int rob(TreeNode root) {
         if(root == null) return 0;
         if(root.left == null && root.right == null) return root.val;
         int maxnum1 = 0;
         int maxnum2 = 0;
         maxnum1 = rob(root.left) + rob(root.right);
         if(root.left != null || root.right != null){
             int num1 = root.left!=null?rob(root.left.left) + rob(root.left.right):0;
             int num2 = root.right!=null?rob(root.right.left) + rob(root.right.right):0;
             maxnum2 = num1 + num2 + root.val;
         }
         return maxnum1>maxnum2?maxnum1:maxnum2;
     }
 }