题解 | #二叉树的镜像#

是递归呀。

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */
/**
 * 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
 */
struct TreeNode* convert(struct TreeNode* proot){
    if(!proot) return NULL;
    struct TreeNode* temp = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    temp->val = proot->val;
    temp->left = convert(proot->right);
    temp->right = convert(proot->left);
    return temp;
}
struct TreeNode* Mirror(struct TreeNode* pRoot ) {
    // write code here
    struct TreeNode* p = pRoot;
    struct TreeNode* resul = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    resul->left = NULL;
    resul->right = NULL;
    resul = convert(p);
    return resul;
}