/**
 * 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 {
    // 1.层序遍历，每次删除两个节点，判断轴对称的两个节点的值是否相等，并将他们的左右子节点的轴对称位置添加进行
    // 2.递归
    public boolean isSymmetric(TreeNode root) {
    //    if (root == null) {
    //     return true;
    //    }
    //    return isMirror(root.left, root.right);
        if (root == null) {
            return true;
        }
        Queue<TreeNode> que = new LinkedList<>();
        que.offer(root.left);
        que.offer(root.right);
        while (!que.isEmpty()) {
            int len = que.size();
            for (int i = len; i > 0; i -= 2) {
                TreeNode tmp1 = que.poll();
                TreeNode tmp2 = que.poll();
                if (tmp1 == null && tmp2 == null) {
                    continue;
                }
                if (tmp1 == null || tmp2 == null || tmp1.val != tmp2.val) {
                    return false;
                }
                que.offer(tmp1.left); //轴对称的添加节点
                que.offer(tmp2.right);
                que.offer(tmp1.right);
                que.offer(tmp2.left);
            }
        }
        return true;
    }

    // public boolean isMirror(TreeNode left, TreeNode right) {
    //     if (left == null && right == null) return true;
    //     if (left == null || right == null) return false;
    //     return left.val == right.val && isMirror(left.left, right.right) && isMirror(left.right, right.left);
    // }
}