Problem-solving idea: swap the nodes of the left and right subtrees, and then call the method recursively.
/**
public 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){
return;
}
//前序遍历每个节点的非叶子节点，交换它的两个子节点，最终得到该二叉树的镜像
TreeNode temp = null;
temp = root.left;
root.left = root.right;
root.right = temp;
//左右子树用递归自己处理就ok了
Mirror(root.left);
Mirror(root.right);
}
}