友情支持
如果您觉得这个笔记对您有所帮助，看在D瓜哥码这么多字的辛苦上，请友情支持一下，D瓜哥感激不尽，😜
|
|
有些打赏的朋友希望可以加个好友，欢迎关注D 瓜哥的微信公众号，这样就可以通过公众号的回复直接给我发信息。
公众号的微信号是: jikerizhi。因为众所周知的原因，有时图片加载不出来。 如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。 |
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.
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.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.解题分析
由于题目给出示例的误导，这里存在一个误区：在不选父节点时，子树选最优解不受任何干扰，也就是说子树的跟节点可选也可以不选。所以，可以将这个问题简化为，分选根节点和不选根节点，两种情况考虑：
-
选父节点，则子树必定不能选根节点；
-
如果不选跟节点，则子树可以在选根节点和不选跟节点中选最优解就好。
然后在这上面两种情况中选出最优解。
参考资料
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 = <font color="red" style="font-family: sans-serif, Arial, Verdana, "Trebuchet MS";">3 + <font color="red" style="font-family: sans-serif, Arial, Verdana, "Trebuchet MS";">3 + <font color="red" style="font-family: sans-serif, Arial, Verdana, "Trebuchet MS";">1 = <b style="font-family: sans-serif, Arial, Verdana, "Trebuchet MS";">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 + <font color="red">5 = 9. 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for House Robber III.
* Memory Usage: 41.3 MB, less than 13.89% of Java online submissions for House Robber III.
*
* Copy from: https://leetcode-cn.com/problems/house-robber-iii/solution/java-2ms-by-horanol/[java 2ms - 打家劫舍 III - 力扣（LeetCode）]
*/
public int rob(TreeNode root) {
int[] robs = doRob(root);
return Math.max(robs[0], robs[1]);
}
public int[] doRob(TreeNode root) {
int[] result = new int[2];
if (Objects.isNull(root)) {
return result;
}
int[] left = doRob(root.left);
int[] right = doRob(root.right);
result[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
result[1] = left[0] + right[0] + root.val;
return result;
}