在上一篇文章中，我们介绍了B-Tree。我们还讨论了search()和traverse()函数。
在这篇文章中，将讨论insert()操作。始终在叶节点处插入一个新密钥。让要插入的密钥为k。像BST一样，我们从根开始向下遍历，直到到达叶节点。到达叶节点后，将密钥插入该叶节点。与BST不同，我们在节点可以包含的键数上有一个预定义的范围。因此，在将密钥插入节点之前，我们确保节点具有额外的空间。
在插入密钥之前，如何确保节点有空间可用于密钥?我们使用一个称为splitChild()的操作，该操作用于拆分节点的子级。请参阅下图以了解拆分。在下图中，x的子y被分为两个节点y和z。请注意，splitChild操作将键向上移动，这就是B树长大的原因，与BST长大的原因不同。
如上所述，要插入新密钥，我们需要从根到叶。在遍历到一个节点之前，我们首先检查该节点是否已满。如果节点已满，我们将其拆分以创建空间。以下是完整的算法。
插入
1)初始化x为根。
2)当x不是叶子时，请执行以下操作
.. a)找到接下来要遍历的x的子代。让孩子成为你。
.. b)如果y未满，则将x更改为指向y。
.. c)如果y已满，请将其拆分并更改x以指向y的两个部分之一。如果k小于y的中键，则将x设置为y的第一部分。 y的其他第二部分。拆分y时，会将键从y移到其父x。
3)当x是叶子时，步骤2中的循环停止。 x必须有空间容纳1个额外的键，因为我们已经预先拆分了所有节点。因此，只需将k插入x。
请注意，该算法遵循Cormen书。它实际上是一种主动插入算法，其中在下到节点之前，如果节点已满则将其拆分。之前进行拆分的好处是，我们永远不会遍历一个节点两次。如果我们在进入节点之前不分割节点，而仅在插入新键(反应性)时才分割节点，则最终可能会再次从叶到根遍历所有节点。在从根到叶的路径上的所有节点都已满的情况下，会发生这种情况。因此，当我们到达叶子节点时，我们将其拆分并向上移动关键点。向上移动键将导致父节点分裂(因为父节点已满)。这种级联效果永远不会在这种主动插入算法中发生。这种主动插入的缺点是，我们可能会进行不必要的拆分。
让我们用最小度为’t’的示例树为3并在初始为空的B树中的整数序列10、20、30、40、50、60、70、80和90理解算法。
最初的root是NULL。让我们首先插入10。 
现在让我们插入20、30、40和50。它们都将被插入根目录，因为节点可以容纳的最大键数是2 * t – 1，即5。
现在让我们插入60。由于根节点已满，因此它将首先拆分为两个，然后将60插入到适当的子节点中。 
现在让我们插入70和80。这些新密钥将被插入到适当的叶子中，而不会进行任何拆分。 
现在让我们插入90。此插入将导致拆分。中键将转到父级。 
以下是上述主动算法的C++实现。
// C++ program for B-Tree insertion
#include
using namespace std;
// 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 the 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, traverse 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)
C[i]->traverse();
cout << " " << keys[i];
}
// Print the subtree rooted with last child
if (leaf == false)
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)
return this;
// If key is not found here and this is a leaf node
if (leaf == true)
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)
{
// 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)
{
// 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)
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)
{
// 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)
{
// 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)
{
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 the 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;
}
// Driver program to test above functions
int main()
{
BTree t(3); // A B-Tree with minium degree 3
t.insert(10);
t.insert(20);
t.insert(5);
t.insert(6);
t.insert(12);
t.insert(30);
t.insert(7);
t.insert(17);
cout << "Traversal of the constucted tree is ";
t.traverse();
int k = 6;
(t.search(k) != NULL)? cout << "\nPresent" : cout << "\nNot Present";
k = 15;
(t.search(k) != NULL)? cout << "\nPresent" : cout << "\nNot Present";
return 0;
}
输出：
Traversal of the constucted tree is 5 6 7 10 12 17 20 30
Present
Not Present参考：
算法入门第三版，作者：Clifford Stein，Thomas H. Cormen，Charles E. Leiserson，Ronald L. Rivest
http://www.cs.utexas.edu/users/djimenez/utsa/cs3343/lecture17.html