题解 | #二叉树的镜像#

 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param pRoot TreeNode类 
     * @return TreeNode类
     */
    TreeNode* Mirror(TreeNode* pRoot) {
        // write code here
//         TreeNode* res;
        if(!pRoot || pRoot->left == nullptr && pRoot->right == nullptr){
            return pRoot;
        }
        // 先完成root的左右子树的交换
        TreeNode* tmp = pRoot->left;
        pRoot->left = pRoot->right;
        pRoot->right = tmp;
        //再进行递归
        pRoot->left = Mirror(pRoot->left);
        pRoot->right = Mirror(pRoot->right);
        return pRoot;
    }
};