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https://github.com/youngyangyang04/leetcode-master.git
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@ -287,24 +287,85 @@ class Solution {
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Python:
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> 动态规划
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```python
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> 暴力递归
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```python3
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def rob(self, root: TreeNode) -> int:
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result = self.robTree(root)
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if root is None:
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return 0
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if root.left is None and root.right is None:
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return root.val
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# 偷父节点
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val1 = root.val
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if root.left:
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val1 += self.rob(root.left.left) + self.rob(root.left.right)
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if root.right:
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val1 += self.rob(root.right.left) + self.rob(root.right.right)
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# 不偷父节点
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val2 = self.rob(root.left) + self.rob(root.right)
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return max(val1, val2)
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```
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> 记忆化递归
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```python3
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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memory = {}
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def rob(self, root: TreeNode) -> int:
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if root is None:
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return 0
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if root.left is None and root.right is None:
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return root.val
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if self.memory.get(root) is not None:
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return self.memory[root]
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# 偷父节点
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val1 = root.val
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if root.left:
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val1 += self.rob(root.left.left) + self.rob(root.left.right)
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if root.right:
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val1 += self.rob(root.right.left) + self.rob(root.right.right)
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# 不偷父节点
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val2 = self.rob(root.left) + self.rob(root.right)
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self.memory[root] = max(val1, val2)
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return max(val1, val2)
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```
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> 动态规划
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```python3
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def rob(self, root: TreeNode) -> int:
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result = self.rob_tree(root)
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return max(result[0], result[1])
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#长度为2的数组,0:不偷,1:偷
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def robTree(self, cur):
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if not cur:
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return (0, 0) #这里返回tuple, 也可以返回list
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left = self.robTree(cur.left)
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right = self.robTree(cur.right)
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#偷cur
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val1 = cur.val + left[0] + right[0]
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#不偷cur
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val2 = max(left[0], left[1]) + max(right[0], right[1])
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return (val2, val1)
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def rob_tree(self, node):
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if node is None:
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return (0, 0) # (偷当前节点金额,不偷当前节点金额)
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left = self.rob_tree(node.left)
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right = self.rob_tree(node.right)
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val1 = node.val + left[1] + right[1] # 偷当前节点,不能偷子节点
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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
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return (val1, val2)
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```
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Go:
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