轻车熟路

方法1：深度优先遍历

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    //用递归实现
    boolean isSymmetrical(TreeNode pRoot)
    {
        if(pRoot==null) return true;

        return isSame(pRoot.left,pRoot.right);
    }

    boolean isSame(TreeNode left,TreeNode right){
        if(left==null&&right==null){
            return true;
        }
        if(left!=null&&right!=null&&left.val==right.val){
            return  isSame(left.left,right.right)&&isSame(left.right,right.left);
        }else{
            return false;
        }


    }
}

方法2：广度优先遍历

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
import java.util.*;
public class Solution {
    boolean isSymmetrical(TreeNode pRoot)
    {
        if(pRoot ==null||(pRoot.left==null&&pRoot.right==null)){
            return true;
        }
        //bfs左子树用
        Queue<TreeNode> queueLeft = new LinkedList<>();
        //bfs右子树用
        Queue<TreeNode> queueRight = new LinkedList<>();
        //左子树根节点入队
        queueLeft.offer(pRoot.left);
        //右子树根节点入队
        queueRight.offer(pRoot.right);

        while(true){
            //左子树与右子树节点数不等，直接false
            if(queueLeft.size()!=queueRight.size()){
                return false;
            }
            //左右子树遍历完毕 返回true
            if(queueLeft.size()==0&&queueRight.size()==0){
                return true;
            }
            //左子树当前层节点数
            int L = queueLeft.size();
            //右子树当前层节点数
            int R = queueRight.size();
            for(int i=0;i<L;i++){

                TreeNode left = queueLeft.poll();
                TreeNode right = queueRight.poll();
                if(left.val!=right.val){
                    //对比对称节点的节点值，不等直接false
                    return false;
                };
                if(left.left!=null&&right.right!=null){
                    //左子树从左向右入队
                    queueLeft.offer(left.left);
                     //右子树从右向左入队
                    queueRight.offer(right.right);
                }else if(left.left!=right.right){
                    //一个为null一个不为null  直接返回false
                    return false;
                }
                //思路同上
                if(left.right!=null&&right.left!=null){
                    queueLeft.offer(left.right);
                    queueRight.offer(right.left);
                }else if(left.right!=right.left){
                    return false;
                }
            }
        }
    }
}