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Merge pull request #335 from z80160280/master
添加 二叉树中递归带着回溯 198 337 python版本
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@ -131,7 +131,20 @@ class Solution {
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```
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Python:
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```python
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class Solution:
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def rob(self, nums: List[int]) -> int:
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if len(nums) == 0:
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return 0
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if len(nums) == 1:
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return nums[0]
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dp = [0] * len(nums)
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dp[0] = nums[0]
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dp[1] = max(nums[0], nums[1])
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for i in range(2, len(nums)):
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dp[i] = max(dp[i-2]+nums[i], dp[i-1])
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return dp[-1]
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```
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Go:
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```Go
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@ -287,6 +287,25 @@ class Solution {
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Python:
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> 动态规划
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```python
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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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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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```
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Go:
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@ -257,6 +257,83 @@ Java:
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Python:
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100.相同的树
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> 递归法
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```python
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class Solution:
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def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
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return self.compare(p, q)
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def compare(self, tree1, tree2):
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if not tree1 and tree2:
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return False
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elif tree1 and not tree2:
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return False
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elif not tree1 and not tree2:
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return True
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elif tree1.val != tree2.val: #注意这里我没有使用else
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return False
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#此时就是:左右节点都不为空,且数值相同的情况
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#此时才做递归,做下一层的判断
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compareLeft = self.compare(tree1.left, tree2.left) #左子树:左、 右子树:左
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compareRight = self.compare(tree1.right, tree2.right) #左子树:右、 右子树:右
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isSame = compareLeft and compareRight #左子树:中、 右子树:中(逻辑处理)
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return isSame
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```
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257.二叉的所有路径
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> 递归中隐藏着回溯
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```python
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class Solution:
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def binaryTreePaths(self, root: TreeNode) -> List[str]:
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result = []
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path = []
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if not root:
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return result
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self.traversal(root, path, result)
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return result
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def traversal(self, cur, path, result):
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path.append(cur.val)
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#这才到了叶子节点
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if not cur.left and not cur.right:
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sPath = ""
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for i in range(len(path)-1):
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sPath += str(path[i])
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sPath += "->"
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sPath += str(path[len(path)-1])
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result.append(sPath)
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return
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if cur.left:
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self.traversal(cur.left, path, result)
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path.pop() #回溯
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if cur.right:
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self.traversal(cur.right, path, result)
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path.pop() #回溯
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```
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> 精简版
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```python
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class Solution:
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def binaryTreePaths(self, root: TreeNode) -> List[str]:
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result = []
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path = ""
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if not root:
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return result
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self.traversal(root, path, result)
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return result
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def traversal(self, cur, path, result):
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path += str(cur.val) #中
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if not cur.left and not cur.right:
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result.append(path)
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return
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if cur.left:
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self.traversal(cur.left, path+"->", result) #左 回溯就隐藏在这里
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if cur.right:
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self.traversal(cur.right, path+"->", result) #右 回溯就隐藏在这里
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```
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Go:
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