小黑隔离日期被多次刷新，深感绝望，得亏阿黄能陪我几天的leetcode之旅:226. 翻转二叉树,257. 二叉树的所有路径

本文主要是介绍小黑隔离日期被多次刷新，深感绝望，得亏阿黄能陪我几天的leetcode之旅:226. 翻转二叉树,257. 二叉树的所有路径，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

226. 翻转二叉树

小黑做法

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:# 递归函数def dfs(node):if not node:return Nonenode.left,node.right = node.right,node.leftdfs(node.left)dfs(node.right)dfs(root)return root

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层次遍历法

# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:if not root:return# 初始化队列q = collections.deque([root])while q:# 根结点出队列top = q.popleft()# 交换操作top.left, top.right = top.right, top.leftif top.left:q.append(top.left)if top.right:q.append(top.right)return root

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257. 二叉树的所有路径

小黑做法

# 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 binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:if not root:return []self.paths = []def dfs(node, path):# 递归出口if not (node.left or node.right):self.paths.append(''.join(path))# 遍历左子树if node.left:path.append('->'+str(node.left.val))dfs(node.left,path)# 左子树结点poppath.pop()# 遍历右子树if node.right:path.append('->'+str(node.right.val))dfs(node.right,path)# 右子树结点poppath.pop()dfs(root,[str(root.val)])return self.paths

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层次遍历

# 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 binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:if not root:return []# 初始化队列q = collections.deque([root])q_str = collections.deque([str(root.val)])paths = []while q:# 出队top = q.popleft()top_str = q_str.popleft()# 输出路径字符串if not (top.left or top.right):paths.append(top_str)continueif top.left:q_str.append(top_str+'->'+str(top.left.val))q.append(top.left)if top.right:q.append(top.right)q_str.append(top_str+'->'+str(top.right.val))return paths