判断二叉树是否镜像

问题：

　　请实现一个函数，用来判断一颗二叉树是不是对称的。注意，如果一个二叉树同此二叉树的镜像是同样的，定义其为对称的。

分析：

　　（1）先序遍历和中序遍历可以唯一确定一棵二叉树

　　（2）左右子树镜像：右子树的遍历规则也需要使用镜像遍历规则

　　（3）分别将先序遍历+中序遍历的结果串接在一起，然后比较左右子树的串接结果。

　　（4）边界情况判断：头结点为空，也属于对称：

code：

 //树节点定义
    public class TreeNode {
        int val = 0;
        TreeNode left = null;
        TreeNode right = null;

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

        }

    }
 //判断方法
    boolean isSymmetrical(TreeNode pRoot) {
        //头结点为空的情况
        if(pRoot==null){
            return true;
        }
      
        StringBuilder leftBuilder1 = new StringBuilder(); //左先序
        StringBuilder leftBuilder2 = new StringBuilder(); //左中序
        //遍历左子树
        Search(pRoot.left,leftBuilder1,leftBuilder2);
        //串接结果
        leftBuilder1.append(leftBuilder2);
        
        StringBuilder rightBuilder1 = new StringBuilder(); //右先序
        StringBuilder rightBuilder2 = new StringBuilder();  //右中序
        //遍历右子树
        mirrorSearch(pRoot.right,rightBuilder1,rightBuilder2);
        //串接结果
        rightBuilder1.append(rightBuilder2);
        //比较判断
        if(leftBuilder1.toString().equals(rightBuilder1.toString())){
            return true;
        }else{
            return false;
        }

    }
//左枝：先序遍历+中序遍历
    public void Search(TreeNode head,StringBuilder preBuilder1,StringBuilder midBuilder){
        if(head!=null){
            preBuilder1.append(head.val);
            if(head.left!=null){
                Search(head.left,preBuilder1,midBuilder);
            }else{
                preBuilder1.append("#");
                midBuilder.append("#");
            }
            midBuilder.append(head.val);
            if(head.right!=null){
                Search(head.right,preBuilder1,midBuilder);
            }else{
                preBuilder1.append("#");
                midBuilder.append("#");
            }
        }
    }
    //右枝：镜像先序遍历+中序遍历
    public void mirrorSearch(TreeNode head,StringBuilder preBuilder1,StringBuilder midBuilder){
        if(head!=null){
            preBuilder1.append(head.val);
            if(head.right!=null){
                mirrorSearch(head.right,preBuilder1,midBuilder);
            }else{
                preBuilder1.append("#");
                midBuilder.append("#");
            }
            midBuilder.append(head.val);
            if(head.left!=null){
                mirrorSearch(head.left,preBuilder1,midBuilder);
            }else{
                preBuilder1.append("#");
                midBuilder.append("#");
            }
        }
    }