From 7d2a897577ff9fefedc377c8e0f1c421c1b879d1 Mon Sep 17 00:00:00 2001
From: haofeng <852172305@qq.com>
Date: Sun, 20 Jun 2021 12:02:07 +0800
Subject: [PATCH] =?UTF-8?q?Update=200337.=E6=89=93=E5=AE=B6=E5=8A=AB?=
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---
 problems/0337.打家劫舍III.md | 89 +++++++++++++++++++++++++++-----
 1 file changed, 75 insertions(+), 14 deletions(-)

diff --git a/problems/0337.打家劫舍III.md b/problems/0337.打家劫舍III.md
index 0a4b752b..3504a574 100644
--- a/problems/0337.打家劫舍III.md
+++ b/problems/0337.打家劫舍III.md
@@ -287,24 +287,85 @@ class Solution {
 
 Python：
 
-> 动态规划
-```python
+> 暴力递归
+
+```python3
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def rob(self, root: TreeNode) -> int:
-        result = self.robTree(root)
+        if root is None:
+            return 0
+        if root.left is None and root.right  is None:
+            return root.val
+        # 偷父节点
+        val1 = root.val
+        if root.left:
+            val1 += self.rob(root.left.left) + self.rob(root.left.right)
+        if root.right:
+            val1 += self.rob(root.right.left) + self.rob(root.right.right)
+        # 不偷父节点
+        val2 = self.rob(root.left) + self.rob(root.right)
+        return max(val1, val2)
+```
+
+> 记忆化递归
+
+```python3
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    memory = {}
+    def rob(self, root: TreeNode) -> int:
+        if root is None:
+            return 0
+        if root.left is None and root.right  is None:
+            return root.val
+        if self.memory.get(root) is not None:
+            return self.memory[root]
+        # 偷父节点
+        val1 = root.val
+        if root.left:
+            val1 += self.rob(root.left.left) + self.rob(root.left.right)
+        if root.right:
+            val1 += self.rob(root.right.left) + self.rob(root.right.right)
+        # 不偷父节点
+        val2 = self.rob(root.left) + self.rob(root.right)
+        self.memory[root] = max(val1, val2)
+        return max(val1, val2)
+```
+
+> 动态规划
+```python3
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def rob(self, root: TreeNode) -> int:
+        result = self.rob_tree(root)
         return max(result[0], result[1])
     
-    #长度为2的数组，0：不偷，1：偷
-    def robTree(self, cur):
-        if not cur:
-            return (0, 0) #这里返回tuple, 也可以返回list
-        left = self.robTree(cur.left)
-        right = self.robTree(cur.right)
-        #偷cur
-        val1 = cur.val + left[0] + right[0]
-        #不偷cur
-        val2 = max(left[0], left[1]) + max(right[0], right[1])
-        return (val2, val1)
+    def rob_tree(self, node):
+        if node is None:
+            return (0, 0) # (偷当前节点金额，不偷当前节点金额)
+        left = self.rob_tree(node.left)
+        right = self.rob_tree(node.right)
+        val1 = node.val + left[1] + right[1] # 偷当前节点，不能偷子节点
+        val2 = max(left[0], left[1]) + max(right[0], right[1]) # 不偷当前节点，可偷可不偷子节点
+        return (val1, val2)
 ```
 
 Go：