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
     */
    void swap(TreeNode* &pRoot1,TreeNode* &pRoot2)
    {
        TreeNode* temp = pRoot1;
        pRoot1 = pRoot2;
        pRoot2 = temp;
    }

    TreeNode* Mirror(TreeNode* pRoot) {
        // write code here
        if(!pRoot)
            return nullptr;
        //左右子树交换
        swap(pRoot->left,pRoot->right);
        //递归左右子树
        pRoot->left = Mirror(pRoot->left);
        pRoot->right = Mirror(pRoot->right);
        return pRoot;
    }
};