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:

Input: [3,2,3,null,3,null,1]

     3
    / \
   2   3
    \   \ 
     3   1

Output: 7 
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
  4   5
  / \   \ 
 1   3   1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/house-robber-iii
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相邻的两个房间如果同时被抢劫会报警，求能抢劫的最大金额。

DFS遍历，每个房间都可以选择抢或者不抢，如果抢的话，那么它两个子节点就不能抢。如果不抢的话，它的两个子节点可抢可不抢，选择最大的。

class Solution {
    public int doRob(TreeNode treeNode) {
        if (treeNode.left == null && treeNode.right == null) {
            return treeNode.val;
        }
        int maxLeft = 0;
        if (treeNode.left != null) {
            maxLeft = noRob(treeNode.left);
        }
        int maxRight = 0;
        if (treeNode.right != null) {
            maxRight = noRob(treeNode.right);
        }
        return treeNode.val + maxLeft + maxRight;
    }
    public int noRob(TreeNode treeNode) {
        if (treeNode.left == null && treeNode.right == null) {
            return 0;
        }
        int maxLeft = 0;
        if (treeNode.left != null) {
            maxLeft = Math.max(noRob(treeNode.left), doRob(treeNode.left));
        }
        int maxRight = 0;
        if (treeNode.right != null) {
            maxRight = Math.max(noRob(treeNode.right), doRob(treeNode.right));
        }
        return maxLeft + maxRight;
    }
    public int rob(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return Math.max(doRob(root), noRob(root));
    }
}