# 337.House-Robber-III

## 题目地址

https://leetcode.com/problems/house-robber-iii/

## 题目描述

``````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.``````

## 代码

### Approach #1 DFS + Memoization

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int rob(TreeNode root) {
return robSub(root, new HashMap<>());
}

private int robSub(TreeNode root, Map<TreeNode, Integer> map) {
if (root == null) return 0;
if (map.containsKey(root)) return map.get(root);

int val = 0;

if (root.left != null) {
val += robSub(root.left.left, map) + robSub(root.left.right, map);
}

if (root.right != null) {
val += robSub(root.right.left, map) + robSub(root.right.right, map);
}

val = Math.max(val + root.val, robSub(root.left, map) + robSub(root.right, map));
map.put(root, val);

return val;
}
}``````

### Approach #2 DFS

1. 如果小偷偷了当前结点，那么它的子结点不能偷

2. 如果小偷不偷当前结点，那么子结点可以偷，也可以不偷（取其中较大的即可）

num[0] is the max value while rob this node, num[1] is max value while not rob this value.

``````public int rob(TreeNode root) {
if (root == null) return 0;
return Math.max(robInclude(root), robExclude(root));
}

public int robInclude(TreeNode node) {
if(node == null) return 0;
return robExclude(node.left) + robExclude(node.right) + node.val;
}

public int robExclude(TreeNode node) {
if(node == null) return 0;
return rob(node.left) + rob(node.right);
}

class Solution {
public int rob(TreeNode root) {
int[] num = dfs(root);
return Math.max(num[0], num[1]);
}

private int[] dfs(TreeNode node) {
if (node == null)        return new int[2];
int[] left = dfs(node.left);
int[] right = dfs(node.right);
int[] res = new int[2];
res[0] = left[1] + right[1] + node.val; // 偷 node, left和right不偷
res[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]); // 不偷 node, 子节点可以偷，也可以不偷
return res;
}
}``````

### Approach #3 Improved Confusion

``````public int rob(TreeNode root) {
int[] res = robSub(root);
return Math.max(res[0], res[1]);
}

private int[] robSub(TreeNode root) {
if (root == null) return new int[2];

int[] left = robSub(root.left);
int[] right = robSub(root.right);
int[] res = new int[2];

res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
res[1] = root.val + left[0] + right[0];

return res;
}``````

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