跳转至

226.invert-binary-tree

Statement

Metadata
  • Link: 翻转二叉树
  • Difficulty: Easy
  • Tag: 深度优先搜索 广度优先搜索 二叉树

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。

 

示例 1:

输入:root = [4,2,7,1,3,6,9]
输出:[4,7,2,9,6,3,1]

示例 2:

输入:root = [2,1,3]
输出:[2,3,1]

示例 3:

输入:root = []
输出:[]

 

提示:

  • 树中节点数目范围在 [0, 100]
  • -100 <= Node.val <= 100

Metadata
  • Link: Invert Binary Tree
  • Difficulty: Easy
  • Tag: Tree Depth-First Search Breadth-First Search Binary Tree

Given the root of a binary tree, invert the tree, and return its root.

 

Example 1:

Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]

Example 2:

Input: root = [2,1,3]
Output: [2,3,1]

Example 3:

Input: root = []
Output: []

 

Constraints:

  • The number of nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100

Solution

# 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 dfs(self, rt: TreeNode) -> None:
        if not rt:
            return

        rt.left, rt.right = rt.right, rt.left
        self.dfs(rt.left)
        self.dfs(rt.right)

    def invertTree(self, root: TreeNode) -> TreeNode:
        self.dfs(root)
        return root

最后更新: October 11, 2023
回到页面顶部