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
        //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;
    }
};