题解 | #二叉树的镜像#
二叉树的镜像
https://www.nowcoder.com/practice/a9d0ecbacef9410ca97463e4a5c83be7
/** * 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 //空树 if (pRoot == nullptr) return nullptr; //只有根节点 if (pRoot->left == nullptr && pRoot->right == nullptr) return pRoot; //后序遍历交换节点 if (pRoot->left != nullptr) Mirror(pRoot->left); if (pRoot->right != nullptr) Mirror(pRoot->right); if (pRoot->left != nullptr && pRoot->right == nullptr) { pRoot->right = pRoot->left; pRoot->left = nullptr; } else if (pRoot->right != nullptr && pRoot->left == nullptr) { pRoot->left = pRoot->right; pRoot->right = nullptr; } else if (pRoot->left == nullptr && pRoot->right == nullptr) { return nullptr; } else { TreeNode* temp = pRoot->left; pRoot->left = pRoot->right; pRoot->right = temp; } return pRoot; } };