LeetCode(226):翻转二叉树-优快云博客

本文链接：https://blog.youkuaiyun.com/ITxiaofeixiang/article/details/110395244

本文探讨了如何通过层次遍历和递归方法解决LeetCode第226题——翻转二叉树。层次遍历采用广度优先策略，而递归方法则是深度优先。两种解法对比，适合提升对树结构操作的理解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

LeetCode(226):翻转二叉树

开头

最近一直再刷树方面的题目，希望尽早掌握！

题意

在这里插入图片描述

题解

一、层次遍历，从上向下翻转

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        //队列实现广度优先遍历
        TreeNode result = root;
        if(root == null){
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(result);
        while(!queue.isEmpty()){
            TreeNode temp = queue.poll();
            TreeNode t1 = temp.left;
            TreeNode t2 = temp.right;
            temp.right = t1;
            temp.left = t2;
            if(t1!=null){
                queue.offer(t1);
            }
            if(t2!=null){
                queue.offer(t2);
            }
        }
        return result;
    }
}

二、递归，从下向上遍历翻转

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        //递归实现深度优先遍历
        if(root == null){
            return root;
        }
        TreeNode l = invertTree(root.left);
        TreeNode r = invertTree(root.right);
        root.left = r;
        root.right = l;
        return root;
    }
}