我正在尝试按级别顺序打印一棵 b 树，但它一直在崩溃。我不确定真正的原因是什么，但我认为它崩溃是因为指针。我正在尝试使用我在网上找到的一个函数，该函数遍历每个级别并将其放入队列中并打印出来，但我遇到了这个问题。如果有人有其他方法，请告诉我。
// C++ program for B-Tree insertion
#include<iostream>
#include <queue>
using namespace std;
int ComparisonCount = 0;
// A BTree node
class BTreeNode
{
int *keys; // An array of keys
int t; // Minimum degree (defines the range for number of keys)
BTreeNode **C; // An array of child pointers
int n; // Current number of keys
bool leaf; // Is true when node is leaf. Otherwise false
public:
BTreeNode(int _t, bool _leaf); // Constructor
// A utility function to insert a new key in the subtree rooted with
// this node. The assumption is, the node must be non-full when this
// function is called
void insertNonFull(int k);
// A utility function to split the child y of this node. i is index of y in
// child array C[]. The Child y must be full when this function is called
void splitChild(int i, BTreeNode *y);
// A function to traverse all nodes in a subtree rooted with this node
void traverse();
// A function to search a key in subtree rooted with this node.
BTreeNode *search(int k); // returns NULL if k is not present.
// Make BTree friend of this so that we can access private members of this
// class in BTree functions
friend class BTree;
};
// A BTree
class BTree
{
BTreeNode *root; // Pointer to root node
int t; // Minimum degree
public:
// Constructor (Initializes tree as empty)
BTree(int _t)
{
root = NULL; t = _t;
}
// function to traverse the tree
void traverse()
{
if (root != NULL) root->traverse();
}
// function to search a key in this tree
BTreeNode* search(int k)
{
return (root == NULL) ? NULL : root->search(k);
}
// The main function that inserts a new key in this B-Tree
void insert(int k);
};
// Constructor for BTreeNode class
BTreeNode::BTreeNode(int t1, bool leaf1)
{
// Copy the given minimum degree and leaf property
t = t1;
leaf = leaf1;
// Allocate memory for maximum number of possible keys
// and child pointers
keys = new int[2 * t - 1];
C = new BTreeNode *[2 * t];
// Initialize the number of keys as 0
n = 0;
}
// Function to traverse all nodes in a subtree rooted with this node
/*void BTreeNode::traverse()
{
// There are n keys and n+1 children, travers through n keys
// and first n children
int i;
for (i = 0; i < n; i++)
{
// If this is not leaf, then before printing key[i],
// traverse the subtree rooted with child C[i].
if (leaf == false)
{
ComparisonCount++;
C[i]->traverse();
}
cout << " " << keys[i];
}
// Print the subtree rooted with last child
if (leaf == false)
{
ComparisonCount++;
C[i]->traverse();
}
}*/
// Function to search key k in subtree rooted with this node
BTreeNode *BTreeNode::search(int k)
{
// Find the first key greater than or equal to k
int i = 0;
while (i < n && k > keys[i])
i++;
// If the found key is equal to k, return this node
if (keys[i] == k)
{
ComparisonCount++;
return this;
}
// If key is not found here and this is a leaf node
if (leaf == true)
{
ComparisonCount++;
return NULL;
}
// Go to the appropriate child
return C[i]->search(k);
}
// The main function that inserts a new key in this B-Tree
void BTree::insert(int k)
{
// If tree is empty
if (root == NULL)
{
ComparisonCount++;
// Allocate memory for root
root = new BTreeNode(t, true);
root->keys[0] = k; // Insert key
root->n = 1; // Update number of keys in root
}
else // If tree is not empty
{
// If root is full, then tree grows in height
if (root->n == 2 * t - 1)
{
ComparisonCount++;
// Allocate memory for new root
BTreeNode *s = new BTreeNode(t, false);
// Make old root as child of new root
s->C[0] = root;
// Split the old root and move 1 key to the new root
s->splitChild(0, root);
// New root has two children now. Decide which of the
// two children is going to have new key
int i = 0;
if (s->keys[0] < k)
{
ComparisonCount++;
i++;
}s->C[i]->insertNonFull(k);
// Change root
root = s;
}
else // If root is not full, call insertNonFull for root
root->insertNonFull(k);
}
}
// A utility function to insert a new key in this node
// The assumption is, the node must be non-full when this
// function is called
void BTreeNode::insertNonFull(int k)
{
// Initialize index as index of rightmost element
int i = n - 1;
// If this is a leaf node
if (leaf == true)
{
ComparisonCount++;
// The following loop does two things
// a) Finds the location of new key to be inserted
// b) Moves all greater keys to one place ahead
while (i >= 0 && keys[i] > k)
{
keys[i + 1] = keys[i];
i--;
}
// Insert the new key at found location
keys[i + 1] = k;
n = n + 1;
}
else // If this node is not leaf
{
// Find the child which is going to have the new key
while (i >= 0 && keys[i] > k)
i--;
// See if the found child is full
if (C[i + 1]->n == 2 * t - 1)
{
ComparisonCount++;
// If the child is full, then split it
splitChild(i + 1, C[i + 1]);
// After split, the middle key of C[i] goes up and
// C[i] is splitted into two. See which of the two
// is going to have the new key
if (keys[i + 1] < k)
i++;
}
C[i + 1]->insertNonFull(k);
}
}
// A utility function to split the child y of this node
// Note that y must be full when this function is called
void BTreeNode::splitChild(int i, BTreeNode *y)
{
// Create a new node which is going to store (t-1) keys
// of y
BTreeNode *z = new BTreeNode(y->t, y->leaf);
z->n = t - 1;
// Copy the last (t-1) keys of y to z
for (int j = 0; j < t - 1; j++)
z->keys[j] = y->keys[j + t];
// Copy the last t children of y to z
if (y->leaf == false)
{
ComparisonCount++;
for (int j = 0; j < t; j++)
z->C[j] = y->C[j + t];
}
// Reduce the number of keys in y
y->n = t - 1;
// Since this node is going to have a new child,
// create space of new child
for (int j = n; j >= i + 1; j--)
C[j + 1] = C[j];
// Link the new child to this node
C[i + 1] = z;
// A key of y will move to this node. Find location of
// new key and move all greater keys one space ahead
for (int j = n - 1; j >= i; j--)
keys[j + 1] = keys[j];
// Copy the middle key of y to this node
keys[i] = y->keys[t - 1];
// Increment count of keys in this node
n = n + 1;
}
void BTreeNode::traverse()
{
std::queue<BTreeNode*> queue;
queue.push(this);
while (!queue.empty())
{
BTreeNode* current = queue.front();
queue.pop();
int i;
for (i = 0; i < n; i++)
{
if (leaf == false)
queue.push(current->C[i]);
cout << " " << current->keys[i] << endl;
}
if (leaf == false)
queue.push(current->C[i]);
}
}
// Driver program to test above functions
int main()
{
BTree t(4); // A B-Tree with minium degree 4
srand(29324);
for (int i = 0; i<200; i++)
{
int p = rand() % 10000;
t.insert(p);
}
cout << "Traversal of the constucted tree is ";
t.traverse();
int k = 6;
(t.search(k) != NULL) ? cout << "\nPresent" : cout << "\nNot Present";
k = 28;
(t.search(k) != NULL) ? cout << "\nPresent" : cout << "\nNot Present";
cout << "There are " << ComparisonCount << " comparison." << endl;
system("pause");
return 0;
}
最佳答案
您的遍历代码使用 this 的字段值就好像它们是 current 的值一样循环体中的节点。
你需要坚持current->像这样在循环体中的成员引用前面(在标有“//*”的行中):
while (!queue.empty())
{
BTreeNode* current = queue.front();
queue.pop();
int i;
for (i = 0; i < current->n; i++) //*
{
if (current->leaf == false) //*
queue.push(current->C[i]);
cout << " " << current->keys[i] << endl;
}
if (current->leaf == false) //*
queue.push(current->C[i]);
}
这是一个强有力的指标，表明所有的东西都符合 current-> 的要求。在现实中想要生活在它所在的功能中this因此不需要明确命名。
与我们在此处获得的大多数调试请求相比，您的代码组织得更好，阅读起来也更愉快，但它仍然相当脆弱，并且包含相当多的臭味，例如 if (current->leaf == false)而不是 if (not current->is_leaf) .
当你把它变成工作状态时，你可能想把它发布到 Code Review 上；我敢肯定，在那里闲逛的经验丰富的编码人员可以为您提供很多关于如何改进代码的宝贵建议。
为了简化原型(prototype)设计和开发，我强烈建议如下:
- 使用
std::vector<>在原型(prototype)阶段代替裸阵列 - 在开发/原型(prototype)制作期间使无效条目无效(将键设置为 -1 并将指针设置为 0)
- 使用
assert()用于记录和检查局部不变量 - 编写准确验证结构不变量的函数，并在每个修改结构的函数之前/之后调用它们
- 用
/Wall /Wextra编译你的代码并清理它，以便它始终在没有警告的情况下进行编译
此外，不要使用 int不分青红皂白；不能变成负数的事物的基本类型是 unsigned (节点度数、当前键数等)。
P.S.:通过固定键数的顺序来构建一致的 B 树会更容易(即对于某些 K，键数可以在 K 和 2*K 之间变化)。将顺序固定在指针的数量上会使事情变得更加困难，一个结果是“顺序”2 的键的数量(其中一个节点允许有 2 到 4 个指针)可以在 1 到 3 之间变化。对于大多数处理 B 树的人们将是一个相当出乎意料的景象!
关于c++ - btree 程序可能因指针而崩溃，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36819023/