Question

这真让我难过。我有一个按城市名称排序的城市二元搜索树。城市还包含人口和GPS坐标。我希望能够通过城市名称或城市坐标从树中删除节点。我按名称删除工作正常，但GPS坐标不起作用。

当我通过GPS删除节点时，我尝试打印二叉树时会出现堆栈溢出。下面是我的一些代码。我无法理解如果我按名称删除它将如何工作，但如果我使用相同的删除方法删除坐标，则无法理解。

我得到的确切错误是＆＃34; EXE中0x013214D6处的未处理异常：0xC00000FD：堆栈溢出（参数：0x00000001,0x00152FFC）。＆＃34;在按坐标删除后，我的打印功能会发生这种情况，但如果我按名称删除则不会。

bool BinaryTree::DeleteByName(string city)
{
    if (GetRoot() != NULL)
    {
        return (DeleteByName(GetRoot(), city));
    }
    return false;
}

TreeNode* BinaryTree::DeleteByName(TreeNode *node, string city)
{
    if (node == NULL)
    {
        return node;
    }
    else if (city < node->Data.name)
    {
        node->Left = DeleteByName(node->Left, city);
    }
    else if (city > node->Data.name)
    {
        node->Right = DeleteByName(node->Right, city);
    }
    else
    {
        if (node->Left == NULL && node->Right == NULL)
        {
            delete node;
            node = NULL;
        }
        else if (node->Left == NULL)        
        {
            TreeNode* temp = node;
            node = node->Right;
            delete temp;
        }
        else if (node->Right == NULL)
        {
            TreeNode* temp = node;
            node = node->Left;
            delete temp;
        }
        else
        {
            cout << "Else";
            TreeNode* temp = MinPtr(node->Right);
            node->Data = temp->Data;
            node->Right = DeleteByName(node->Right, temp->Data.name);
        }
    }
    return node;
}

bool BinaryTree::DeleteByCoord(pair<double, double> coords)
{
    if (GetRoot() == NULL)
    {
        return false;
    }
    else
    {
        return DeleteByCoord(GetRoot(), coords);
    }
}

bool BinaryTree::DeleteByCoord(TreeNode* node, pair<double, double> coords)
{
    bool result;

    if (node == NULL)
    {
        return false;
    }
    else
    {
        if (node->Data.coordinates.first == coords.first && node->Data.coordinates.second == coords.second)
        {
            return (DeleteByName(node, node->Data.name));           
        }

        result = DeleteByCoord(node->Left, coords);
        if (result == true)
        {
            return result;
        }

        return DeleteByCoord(node->Right, coords);
    }
}




void BinaryTree::Insert(City city)
{
    TreeNode* temp = new TreeNode(city);
    if (GetRoot() == NULL)
    {
        root = temp;
    }
    else
    {
        Insert(temp, GetRoot());
    }
}

void BinaryTree::Insert(TreeNode* toAdd, TreeNode* addHere)
{
    if (toAdd->Data < addHere->Data)
    {
        if (addHere->Left != NULL)
        {
            Insert(toAdd, addHere->Left); 
        }
        else
        {
            addHere->Left = toAdd;
        }
    }
    else if (toAdd->Data > addHere->Data)
    {
        if (addHere->Right != NULL)
        {
            Insert(toAdd, addHere->Right);
        }
        else
        {
            addHere->Right = toAdd;
        }
    }
}

void BinaryTree::InOrderTraversal(TreeNode* node)
{
    if (node != NULL)
    {
        InOrderTraversal(node->Left);
        cout << node->Data << endl;
        InOrderTraversal(node->Right);
    }
}

void BinaryTree::InOrderTraversal()
{
    InOrderTraversal(GetRoot());
}

TreeNode* BinaryTree::GetRoot()
{
    return root;
}

TreeNode* BinaryTree::MinPtr(TreeNode* node)
{
    while (node->Left != NULL)
    {
        node = node->Left;
    }   
    return node;
}

Answer 1

删除节点时，还需要更新指向已删除节点的父指针。请注意：

当你直接调用DeleteByName时，它搜索所需的节点并返回自动设置为父节点指针的NULL指针：

else if (city < node->Data.name)
{
    node->Left = DeleteByName(node->Left, city);
}
else if (city > node->Data.name)
{
    node->Right = DeleteByName(node->Right, city);
}

但是当您从coordinates方法调用DeleteByName时，您不会重置父{q}个Left个Right指针：

if (node->Data.coordinates.first == coords.first && node->Data.coordinates.second == coords.second)
{
    return (DeleteByName(node, node->Data.name));           
}

反过来DeleteByName已经收到所需的节点，它不执行递归调用，也不重置父指针：

else
{
    if (node->Left == NULL && node->Right == NULL)
    {
        delete node;
        node = NULL;
    }
    //... same here
}

注意：您的代码中还有许多问题。有些引人注目：

DeleteByName返回指针，但DeleteByCoord返回bool，您在DeleteByCoord
避免直接比较双打，比较结果可能是错误的。有关详细信息，请参阅question和explanation。

从BST删除节点时出错

1 个答案: