Merge pull request #422 from haofengsiji/update

Update 0337.打家劫舍III.md
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程序员Carl
2021-06-23 09:38:07 +08:00
committed by GitHub

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@ -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
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)
```
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