题目：
操作给定的二叉树，将其变换为源二叉树的镜像。
思路：
前序遍历这棵树的每个节点，如果遍历到的节点有子节点，就交换它的两个子节点。当交换完所有非叶节点的左右子节点之后，就得到了树的镜像。
过去的代码： 
class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

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

    }
}
public class Solution {
    public void Mirror(TreeNode root) {         if(root==null||(root.left==null&&root.right==null)) {             return;         }         TreeNode temp=root.left;         root.left=root.right;         root.right=temp;         if(root.left!=null) {             Mirror(root.left);         }         if(root.right!=null) {             Mirror(root.right);         }     }
}
再次写的代码，带测试用例： 
class TreeNode {     TreeNode left;     TreeNode right;     int val;     TreeNode(int val) {         this.val=val;     }
}
public class Solution01 {     public void Mirror(TreeNode root) {         if (root==null||(root.left==null&&root.right==null)) {             return;         }         TreeNode temp=root.left;         root.left=root.right;         root.right=temp;         Mirror(root.left);         Mirror(root.right);     }     public static void main(String[] args) {         TreeNode t1=new TreeNode(8);         TreeNode t2=new TreeNode(6);         TreeNode t3=new TreeNode(10);         TreeNode t4=new TreeNode(5);         TreeNode t5=new TreeNode(7);         TreeNode t6=new TreeNode(9);         TreeNode t7=new TreeNode(11);         t1.left=t2;         t1.right=t3;         t2.left=t4;         t2.right=t5;         t3.left=t6;         t3.right=t7;         Solution01 s=new Solution01();         s.Mirror(t1);         System.out.println(t1.left.right.val);     }
}