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

输入：root = [2,1,3]
输出：[2,3,1]

输入：root = []
输出：[]

import java.util.LinkedList;
import java.util.Queue;
/**
 * 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 {
    /*
    // 法一:BFS
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return root;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            while (levelSize-- > 0) {
                TreeNode poll = queue.poll();
                swap(poll);
                if (poll.left != null) {
                    queue.add(poll.left);
                }
                if (poll.right != null) {
                    queue.add(poll.right);
                }
            }
        }
        return root;
    }
    */
    /**
     * 方法二：
      前后序遍历都可以，中序不行，因为先左孩子交换孩子，再根交换孩子（做完后，右孩子已经变成了原来的左孩子），再右孩子交换孩子（此时其实是对原来的左孩子做交换）
     */
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        invertTree(root.left);
        invertTree(root.right);
        swap(root);
        return root;
    }
        //交换左、右孩子节点
        public void swap(TreeNode root) {
            TreeNode temp = root.left;
            root.left = root.right;
            root.right = temp;
        }
}