Question

所以即时尝试实现Order 2 B-Tree但是我对编程很陌生，特别是在c ++中我为每个节点创建了这个结构

    struct BTreeNode
{
    int data[2];
    BTreeNode **ChildLow;
    BTreeNode **ChildMid;
    BTreeNode **ChildHigh;

};

因此，当我尝试将下一个节点设置为搜索孩子时，我不断收到它想要的BTreeNode类型的编译错误，它不是吗？

bool search(BTreeNode *root, int value)
{
        BTreeNode currentNode = *root;
        bool found = false;
        bool searchComplete = false;
        while(found == false || searchComplete == false)
        {
            if (currentNode == NULL)
            {
                searchComplete == true;
                return false;
            }
            if (currentNode.data[1] == value)
            {
                found == true;
            }
            else if(value > currentNode.data[1])
            {
                if (currentNode.data[2] == value)
                {
                    found == true;
                }
                else if(value > currentNode.data[2])
                {
                    currentNode == currentNode.ChildHigh;

                }
                else
                {
                    currentNode == currentNode.ChildMid;
                }
            }
            else
            {
                currentNode == currentNode.ChildLow;
            }
        }


}

当我将其与null进行比较时，它也会显示错误。

以下是错误：

1   IntelliSense: no operator "==" matches these operands
        operand types are: BTreeNode == int
2   IntelliSense: no operator "==" matches these operands
        operand types are: BTreeNode == BTreeNode **    
3   IntelliSense: no operator "==" matches these operands
        operand types are: BTreeNode == BTreeNode **    
4   IntelliSense: no operator "==" matches these operands
        operand types are: BTreeNode == BTreeNode **

1是Null错误，其余是指向孩子的指针

任何帮助将不胜感激

谢谢

Answer 1

你得到的第一个错误是因为currentNode是BTreeNode而不是指针，应该与NULL进行比较。可能的解决方法是：

BTreeNode* currentNode = root;
/*...code...*/
if (currentNode == NULL)
/*...code...*/
if (currentNode->data[1] == value)
/*...code...*/
currentNode = currentNode->ChildHigh
/*...code...*/
currentNode = currentNode->ChildMid
/*...code...*/
currentNode = currentNode->ChildLow

另请注意，当您输入：

时

//searchComplete == true; this compares
searchComplete = true; // this assigns

和

//found == true; this compares
found = true; this assigns

您没有为searchComplete指定true或找不到，您实际上正在进行比较。

修改

此外，如果您的节点应该指向高，中，低的孩子，那么您不应该使用＆＃39; **＆＃39;，您应该写

BTreeNode* childHigh; BTreeNode* childMid; BTreeNode* childLow;

这意味着他们指向你的孩子。

Answer 2

TL; DR

试试这个，然后找出差异。

struct BTreeNode
{
    int data[2];
    BTreeNode *ChildLow;
    BTreeNode *ChildMid;
    BTreeNode *ChildHigh;

};

bool search(BTreeNode *node, int value)
{
    while(node != NULL)
    {
        if (node->data[0] == value || node->data[1] == value) {
            return true;      // found!
        }

        // search in a subtree
        if(value > node->data[1]) {
            node = node->ChildHigh;
        } else if(value > node->data[0]) {
            node = node->ChildMid;
        } else {
            node = node->ChildLow;
        }
    }

    return false;  // at the bottom of the tree - not found
}

B-Tree搜索引用子数据类型问题

2 个答案: