我在二进制搜索树中有一个删除节点功能，如下所示：
struct Node *_removeNode(struct Node *curr, TYPE val) {
/* FIX ME */
printf("curr->val: %.1f...%p\t val: %.1f\n", curr->val, curr, val);
struct Node* to_del = NULL;
if(val < curr->val){
return _removeNode(curr->left, val);
} else if( val > curr->val) {
return _removeNode(curr->right, val);
} else {
//case 1 -> curr is a leaf
if(curr->left == NULL && curr->right == NULL){
printf("removed %.1f\n", curr->val);
to_del = curr;
free(to_del);
}
//case if left node is not null but right is
if(curr->left != NULL && curr->right == NULL){
to_del = curr;
printf("to_del: %p...%.1f\t", to_del, to_del->val);
curr = curr->left;
printf("new curr: %p...%.1f\n", curr, curr->val);
free(to_del);
}
//case if right node is not null but left is
if(curr->left == NULL && curr->right != NULL){
to_del = curr;
printf("to_del: %p...%.1f\t", to_del, to_del->val);
curr = curr->right;
printf("new curr: %p...%.1f\n", curr, curr->val);
free(to_del);
}
//case both are not not null
if(curr->left != NULL && curr->right != NULL){
to_del = curr;
printf("to_del: %p...%.1f\t", to_del, to_del->val);
curr->right->left = curr->left;
curr = curr->right;
printf("new curr: %p...%.1f\n", curr, curr->val);
free(to_del);
}
}
to_del = NULL;
return to_del;
}
该代码应该返回指向NULL的指针以通过测试用例，但是我尝试的所有操作都会引发内存错误。有没有一种方法可以将结构设置为NULL然后返回它？
最佳答案
看着：
if(curr->left == NULL && curr->right == NULL){
to_del = curr;
free(to_del);
}
if(curr->left != NULL && curr->right == NULL){
如果第一个
if为true，则将释放curr指向的内存。当您在第二个if curr->left中执行操作时，您将收到一个错误消息，因为不再有->left。也许您应该使用
else if？关于c - 释放结构指针，然后将其设置为null，然后返回引发错误，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/33598205/