力扣热题100之翻转二叉树

题目

给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。

代码

方法一：递归

# 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        cur=self.invertTree(root.left)root.left=self.invertTree(root.right)root.right=curreturn root 

方法二：栈迭代

# 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        stack=[root]while stack:node=stack.pop()if node:stack.append(node.left)stack.append(node.right)node.left,node.right=node.right,node.leftreturn root

方法三：队迭代

# 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        queue=deque([root])while queue:for _ in range(len(queue)):node=queue.popleft()if node.left:queue.append(node.left)if node.right:queue.append(node.right)node.left,node.right=node.right,node.leftreturn root

这道题的方法和求二叉树的深度方法一致