226. 翻转二叉树 - 力扣（LeetCode）

方法1：递归法

• 时间：O(n) ------ 每个节点访问一次
• 空间：O(h) ------ 递归栈深度（h 为树高，最坏 O(n)，平均 O(log n)）

# encoding = utf-8
# 开发者：Alen
# 开发时间： 12:24 
# "Stay hungry，stay foolish."

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def invertTree(self, root):
        """
        :type root: Optional[TreeNode]
        :rtype: Optional[TreeNode]
        """
        if not root:
            return None

        root.left, root.right = root.right, root.left

        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

方法2：迭代法（使用栈）

• 时间：O(n)
• 空间：O(h) ------ 栈最多存 h 个节点

# encoding = utf-8
# 开发者：Alen
# 开发时间： 12:24 
# "Stay hungry，stay foolish."

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def invertTree(self, root):
        """
        :type root: Optional[TreeNode]
        :rtype: Optional[TreeNode]
        """
        if not root:
            return None

        stack = [root]
        while stack:
            node = stack.pop()

            node.left, node.right = node.right, node.left

            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
        return root

结果

解题思路