c - 释放结构体指针，然后将其设置为 null，然后返回抛出错误

Question

我在二叉搜索树中有一个删除节点函数，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;

}

代码应该返回一个指向 NULL 的指针，以便通过测试用例，但我尝试的所有操作都会引发内存错误。有没有办法可以将结构设置为 NULL 然后返回它？

Answer 1

看:

    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为真，则curr指向的内存将被释放。然后，当您在第二个 if curr->left 中执行操作时，您会收到错误，因为不再有 ->left 了。

也许你应该使用else if？