LeetCode：226翻转二叉树

_日拱一卒2026-05-01 13:08

方法一：递归法

java 复制代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null){
            return null;
        }
        //左右节点交换
        TreeNode temp = root.right;
        root.right = root.left;
        root.left = temp;
        //递归左右节点
        invertTree(root.left);
        invertTree(root.right);

        return root;
    }
}

方法二：

java 复制代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
 class Solution{
    public TreeNode invertTree(TreeNode root){
        if(root == null){
            return null;
        }

        Queue<TreeNode> queue = new LinkedList<>();

        queue.offer(root);

        while(!queue.isEmpty()){
            TreeNode current = queue.poll();
            //交换左右节点
            TreeNode temp = current.left;
            current.left = current.right;
            current.right = temp;

            //左右节点不为空则入队
            if(current.left != null){
                queue.offer(current.left);
            }
            if(current.right != null){
                queue.offer(current.right);
            }
        } 
        return root;
    }
 }

迭代法思路：

先将根节点入队

while循环，只要队列不为空就交换当前要出队节点的左右节点，然后判断当前出队节点的下面是否还有左右节点，如果有就继续入队，循环。

爆栈，层序遍历，逐层打印。寻找最短路径使用队列；核心：new Queue -> offer(root) -> while(!isEmpty) -> poll() -> 处理逻辑 -> offer(children)