337 House Robber III

Author: int32bit

README.md

@@ -138,6 +138,7 @@
 + [237 Delete Node in a Linked List(O(1)删除单链表节点)](algorithms/DeleteNodeinaLinkedList)
 + [238 Product of Array Except Self(数组操作)](algorithms/ProductofArrayExceptSelf)
 + [242 Valid Anagram(hash表，判断两个字符串出现的字符和次数均相同)](algorithms/ValidAnagram)
++ [337 House Robber III(DFS，递归)](algorithms/HouseRobberIII)
 + [338 Counting Bits(二进制1的个数，迭代法)](algorithms/CountingBits)
 
 ## Database

algorithms/HouseRobberIII/READMD.md

@@ -0,0 +1,68 @@
+## 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`.
+
+## Solution
+
+最开始想到直接层次遍历，获取每一层的值，比如Example 2为`[3,9,5]`,然后转化为[House Robber](../HouseRobber)求解。时间复杂度时O(n),空间复杂度是O(n)。
+
+看提示说使用DFS，一个节点要么选择要么不选择，设当前节点为p，yes为选择p节点的值，no为不选择p节点的值,f(p)为p节点的值，则：
+
+* 若p为null，则yes = 0， no = 0;
+* 若选择p，则不能选择p的孩子节点，则`p->yes = f(p->left->no) + f(p->right->no) + p->val`;
+* 若不选择p节点，则结果为孩子节点的最大值，即`p->no = max(f(p->left->yes), f(p->left->no)) + max(f(p->right->yes), f(p->right->no))`
+
+```cpp
+void dfs(TreeNode *root, int *yes, int *no) {
+	if (root == nullptr) {
+		*yes = 0;
+		*no = 0;
+		return;
+	}
+	int left_yes, left_no, right_yes, right_no;
+	dfs(root->left, &left_yes, &left_no);
+	dfs(root->right, &right_yes, &right_no);
+	*yes = left_no + right_no + root->val;
+	*no = max(left_yes, left_no) + max(right_yes, right_no);
+}
+```
+
+最后返回`max(root->yes, root->no)`即可。
+
+```cpp
+int rob(TreeNode *root) {
+	int yes = 0, no = 0;
+	dfs(root, &yes, &no);
+	return max(yes, no);
+}
+```
+
+## 参考
+
+1. [House Robber](../HouseRobber)
+2. [House Robber II](../HouseRobberII)
+3. [House Robber III](../HouseRobberIII)

algorithms/HouseRobberIII/tree.cpp

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+#include <iostream>
+#include <vector>
+#include <string>
+#include <algorithm>
+#include <cstdio>
+using namespace std;
+struct TreeNode {
+	int val;
+	TreeNode *left;
+	TreeNode *right;
+	TreeNode(int x) : val(x), left(nullptr), right(nullptr){}
+};
+class Solution {
+public:
+	int rob(TreeNode *root) {
+		int yes = 0, no = 0;
+		dfs(root, &yes, &no);
+		return max(yes, no);
+	}
+private:
+	void dfs(TreeNode *root, int *yes, int *no) {
+		if (root == nullptr) {
+			*yes = 0;
+			*no = 0;
+			return;
+		}
+		int left_yes, left_no, right_yes, right_no;
+		dfs(root->left, &left_yes, &left_no);
+		dfs(root->right, &right_yes, &right_no);
+		*yes = left_no + right_no + root->val;
+		*no = max(left_yes, left_no) + max(right_yes, right_no);
+	}
+};
+TreeNode *mk_node(int val)
+{
+	return new TreeNode(val);
+}
+TreeNode *mk_child(TreeNode *root, TreeNode *left, TreeNode *right)
+{
+	root->left = left;
+	root->right = right;
+	return root;
+}
+TreeNode *mk_child(TreeNode *root, int left, int right)
+{
+	return mk_child(root, new TreeNode(left), new TreeNode(right));
+}
+TreeNode *mk_child(int root, int left, int right)
+{
+	return mk_child(new TreeNode(root), new TreeNode(left), new TreeNode(right));
+}
+int main(int argc, char **argv)
+{
+	Solution solution;
+	TreeNode *root = mk_child(4, 1, 0);
+	return 0;
+}