226. Invert Binary Tree

最新推荐文章于 2025-05-24 17:38:15 发布

EchoNUIG

最新推荐文章于 2025-05-24 17:38:15 发布

阅读量56

点赞数

文章标签： leetcode 算法职场和发展

本文链接：https://blog.youkuaiyun.com/weixin_57706549/article/details/127975426

版权

本文介绍了一种翻转二叉树的算法实现，通过递归方式分别翻转左右子树，并交换根节点的左右子节点，最终完成整棵树的翻转。文中详细解释了递归逻辑及核心代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

class Solution {
    public TreeNode invertTree(TreeNode root) {
     if(root == null) return null;
     invertTree(root.left);
     invertTree(root.right);
     reverseChild(root);
     return root;

}
    public void reverseChild(TreeNode root){
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
    }
}

Hints:

1: if(root == null) return null; It is important to focus on the type of retrun, if the key word is "void", we just use "return", or if it is "String or TreeNode", we should return "null"

2: Understand the theory of recursion, the logic and the code of inverting Binary Tree.