226. 翻转二叉树

226. 翻转二叉树https://leetcode.cn/problems/invert-binary-tree/
难度简单1361

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] 内
• -100 <= Node.val <= 100

通过次数501,076提交次数631,866

/**
 * 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) {

        //递归翻转

        if(root==null) return root;
        TreeNode node= root.right;
        root.right = root.left;
        root.left = node;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}

/**
 * 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) {
         //广度优先遍历 + 右->左
        
        if(root==null) return root;

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while(!queue.isEmpty())
        {
            int size = queue.size();
            while(size>0)
            {
                TreeNode node = queue.poll();
                TreeNode temp = node.right;
                node.right = node.left;
                node.left = temp;
                if(node.right!=null) queue.offer(node.right);
                if(node.left!=null) queue.offer(node.left);
                size--;
            }
        }
        return root;

    }
}