Question

我的二叉树删除功能有问题，特别是两个子案例。我的删除代码如下所示，名为with-in

的相关函数

treeNode* delete(treeNode* root, treeNode* del)
{
    treeNode *tempLeft; //node left of curr
    treeNode *tempRight; //node right of curr
    treeNode *save_node;
    treeNode *par;

    par = parent(root, del); // Get the parent of this node.

  // Take care of the case where the node to delete is a leaf node.
    if (isLeaf(del))
    {
        // Deleting the only node in the tree.
        if (par == NULL)
        {
            free(root); // free the memory for the node.
            return NULL;
        }
        // Deletes the node if it's a left child.
        if (compareDates(par->people, del->people) > 0)
        {
            free(par->left); // Free the memory for the node.
            par->left = NULL;
        }
        // Deletes the node if it's a right child.
        else
        {
            free(par->right); // Free the memory for the node.
            par->right = NULL;
        }
        return root; // Return the root of the new tree.
    }
    if (hasOnlyLeftChild(del))
    {

        // Deleting the root node of the tree.
        if (par == NULL)
        {
            save_node = del->left;
            save_node->top = 1;
            free(del); // Free the node to delete.
            return save_node; // Return the new root node of the resulting tree.
        }

        // Deletes the node if it's a left child.
        if (compareDates(par->people, del->people) > 0)
        {
            save_node = par->left; // Save the node to delete.
            par->left = save_node->left; // Readjust the parent pointer.
            free(save_node); // Free the memory for the deleted node.
        }

        // Deletes the node if it's a right child.
        else
        {
            save_node = par->right; // Save the node to delete.
            par->right = save_node->left; // Readjust the parent pointer.
            free(save_node); // Free the memory for the deleted node.
        }

        return root; // Return the root of the tree after the deletion.
    }
        // Takes care of the case where the deleted node only has a right child.
    if (hasOnlyRightChild(del))
    {
        // Node to delete is the root node.
        if (par == NULL)
        {
            save_node = del->right;
            save_node->top = 1;
            free(del);
            return save_node;
        }
        // Delete's the node if it is a left child.
        if (compareDates(par->people, del->people) > 0)
        {
            save_node = par->left;
            par->left = save_node->right;
            free(save_node);
        }
        // Delete's the node if it is a right child.
        else
        {
            save_node = par->right;
            par->right = save_node->right;
            free(save_node);
        }
        return root;
    }
    if(hasTwoChildren(del))
    {
            treeNode *temp;

            save_node = del->left;
            save_node = maxVal(save_node);
            temp = save_node;
            delete(root,save_node);
            del->people = temp->people;
            free(temp);
    }
}

使用两个子案例我在左子树副本中找到最大值到temp var然后在save_node上调用delete然后用temps数据覆盖del中的数据，这样我就不必担心删除一个有两个孩子的节点并修复指针。在我看来应该工作。我遇到的问题是，当我在save_node上运行delete时，它也正在删除我创建的temp，所以当我去复制temp-＆gt; people in del-＆gt; people时，它的NULL为NULL但是我从不触摸temp delete在save_node上运行。

这是一个记忆错误吗，我是自由还是写到记忆中我没有让我得到这个奇怪的结果？是我的删除代码是正确的，我导致错误发生在我的代码的另一部分的问题吗？我是如此坚持这样一个事实：我认为逻辑是对的，我不能让自己去看问题

Answer 1

如果我理解正确，temp和save_node在有两个孩子的情况下指向相同的地址。因此，当您稍后释放此地址时，temp和save_node都会被释放。

二进制删除两个子案例

1 个答案: