思路

• 判断二叉树是否镜像
• 抓住镜像的条件进行BFS
• 镜像及左子树根节点.val == 右子树根节点.val
• 然后递归处理

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

    private boolean dfs(TreeNode left,TreeNode right){
        if(left == null || right == null)
            return left == null && right == null;
        return left.val == right.val && dfs(left.left,right.right) && dfs(left.right,right.left);
    }
}