原题出处：https://leetcode.cn/leetbook/read/top-interview-questions-easy/xn7ihv/

解法一：

/**
 * 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 boolean isSymmetric(TreeNode root) {
        return isCheck(root.left,root.right);
    }

    public boolean isCheck(TreeNode left,TreeNode right) {
        if (left == null && right == null) {
            return true;
        }

        if (left == null || right == null || left.val != right.val) {
            return false;
        }

        return isCheck(left.left,right.right) && isCheck(left.right,right.left);
    }
}

思路：校验对称二叉树，就是需要校验左子树和右子树的值相等，递归校验即可。

解法二：

/**
 * 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 boolean isSymmetric(TreeNode root) {

        Queue<TreeNode> queue = new LinkedList();
        queue.add(root.left);
        queue.add(root.right);

        while (!queue.isEmpty()) {
            TreeNode left = queue.poll(),right = queue.poll();
            if (left == null && right == null) {
                continue;
            }
            if (left == null ^ right == null || left.val != right.val) {
                return false;
            }
            queue.add(left.left);
            queue.add(right.right);
            queue.add(left.right);
            queue.add(right.left);
        }

        return true;

    }


}

思路：非递归的方式，也是将二叉树的左子树与右子树进行比较即可。就是要注意将节点入队列的时候要注意顺序。