题解 | #二叉树的镜像#C++二叉树镜像暴力递归解法

/**
 * 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
     */
    void answer(TreeNode*root)
    {
        if(root)
        {
            TreeNode*t=root->left;
            root->left=root->right;
            root->right=t;
            answer(root->left);
            answer(root->right);
        }
    }
    TreeNode* Mirror(TreeNode* pRoot) 
    {
        answer(pRoot);
        return pRoot;
    }
};