Deletion from a B-tree
In this tutorial, you will learn how to delete a key from a b-tree. Also, you will find working examples of deleting keys from a B-tree in C, C++, Java and Python.
Deleting an element on a B-tree consists of three main events: searching the node where the key to be deleted exists, deleting the key and balancing the tree if required.
While deleting a tree, a condition called underflow may occur. Underflow occurs when a node contains less than the minimum number of keys it should hold.
The terms to be understood before studying deletion operation are:
- Inorder Predecessor
The largest key on the left child of a node is called its inorder predecessor. - Inorder Successor
The smallest key on the right child of a node is called its inorder successor.
Deletion Operation
Before going through the steps below, one must know these facts about a B tree of degree m.
- A node can have a maximum of m children. (i.e. 3)
- A node can contain a maximum of
m - 1keys. (i.e. 2) - A node should have a minimum of
⌈m/2⌉children. (i.e. 2) - A node (except root node) should contain a minimum of
⌈m/2⌉ - 1keys. (i.e. 1)
There are three main cases for deletion operation in a B tree.
Case I
The key to be deleted lies in the leaf. There are two cases for it.
- The deletion of the key does not violate the property of the minimum number of keys a node should hold.In the tree below, deleting 32 does not violate the above properties.
Deleting a leaf key (32) from B-tree - The deletion of the key violates the property of the minimum number of keys a node should hold. In this case, we borrow a key from its immediate neighboring sibling node in the order of left to right.First, visit the immediate left sibling. If the left sibling node has more than a minimum number of keys, then borrow a key from this node.
Else, check to borrow from the immediate right sibling node.
In the tree below, deleting 31 results in the above condition. Let us borrow a key from the left sibling node.
Deleting a leaf key (31) If both the immediate sibling nodes already have a minimum number of keys, then merge the node with either the left sibling node or the right sibling node. This merging is done through the parent node.
Deleting 30 results in the above case.
Delete a leaf key (30)
Case II
If the key to be deleted lies in the internal node, the following cases occur.
- The internal node, which is deleted, is replaced by an inorder predecessor if the left child has more than the minimum number of keys.
Deleting an internal node (33) - The internal node, which is deleted, is replaced by an inorder successor if the right child has more than the minimum number of keys.
- If either child has exactly a minimum number of keys then, merge the left and the right children.
Deleting an internal node (30) After merging if the parent node has less than the minimum number of keys then, look for the siblings as in Case I.
Case III
In this case, the height of the tree shrinks. If the target key lies in an internal node, and the deletion of the key leads to a fewer number of keys in the node (i.e. less than the minimum required), then look for the inorder predecessor and the inorder successor. If both the children contain a minimum number of keys then, borrowing cannot take place. This leads to Case II(3) i.e. merging the children.
Again, look for the sibling to borrow a key. But, if the sibling also has only a minimum number of keys then, merge the node with the sibling along with the parent. Arrange the children accordingly (increasing order).
Python Examples
/* Deleting a key on a B-tree in Python */
/* Btree node */
class BTreeNode:
def __init__(self, leaf=False):
self.leaf = leaf
self.keys = []
self.child = []
class BTree:
def __init__(self, t):
self.root = BTreeNode(True)
self.t = t
/* Insert a key */
def insert(self, k):
root = self.root
if len(root.keys) == (2 * self.t) - 1:
temp = BTreeNode()
self.root = temp
temp.child.insert(0, root)
self.split_child(temp, 0)
self.insert_non_full(temp, k)
else:
self.insert_non_full(root, k)
/* Insert non full */
def insert_non_full(self, x, k):
i = len(x.keys) - 1
if x.leaf:
x.keys.append((None, None))
while i >= 0 and k[0] < x.keys[i][0]:
x.keys[i + 1] = x.keys[i]
i -= 1
x.keys[i + 1] = k
else:
while i >= 0 and k[0] < x.keys[i][0]:
i -= 1
i += 1
if len(x.child[i].keys) == (2 * self.t) - 1:
self.split_child(x, i)
if k[0] > x.keys[i][0]:
i += 1
self.insert_non_full(x.child[i], k)
/* Split the child */
def split_child(self, x, i):
t = self.t
y = x.child[i]
z = BTreeNode(y.leaf)
x.child.insert(i + 1, z)
x.keys.insert(i, y.keys[t - 1])
z.keys = y.keys[t: (2 * t) - 1]
y.keys = y.keys[0: t - 1]
if not y.leaf:
z.child = y.child[t: 2 * t]
y.child = y.child[0: t - 1]
/* Delete a node */
def delete(self, x, k):
t = self.t
i = 0
while i < len(x.keys) and k[0] > x.keys[i][0]:
i += 1
if x.leaf:
if i < len(x.keys) and x.keys[i][0] == k[0]:
x.keys.pop(i)
return
return
if i < len(x.keys) and x.keys[i][0] == k[0]:
return self.delete_internal_node(x, k, i)
elif len(x.child[i].keys) >= t:
self.delete(x.child[i], k)
else:
if i != 0 and i + 2 < len(x.child):
if len(x.child[i - 1].keys) >= t:
self.delete_sibling(x, i, i - 1)
elif len(x.child[i + 1].keys) >= t:
self.delete_sibling(x, i, i + 1)
else:
self.delete_merge(x, i, i + 1)
elif i == 0:
if len(x.child[i + 1].keys) >= t:
self.delete_sibling(x, i, i + 1)
else:
self.delete_merge(x, i, i + 1)
elif i + 1 == len(x.child):
if len(x.child[i - 1].keys) >= t:
self.delete_sibling(x, i, i - 1)
else:
self.delete_merge(x, i, i - 1)
self.delete(x.child[i], k)
/* Delete internal node */
def delete_internal_node(self, x, k, i):
t = self.t
if x.leaf:
if x.keys[i][0] == k[0]:
x.keys.pop(i)
return
return
if len(x.child[i].keys) >= t:
x.keys[i] = self.delete_predecessor(x.child[i])
return
elif len(x.child[i + 1].keys) >= t:
x.keys[i] = self.delete_successor(x.child[i + 1])
return
else:
self.delete_merge(x, i, i + 1)
self.delete_internal_node(x.child[i], k, self.t - 1)
/* Delete the predecessor */
def delete_predecessor(self, x):
if x.leaf:
return x.pop()
n = len(x.keys) - 1
if len(x.child[n].keys) >= self.t:
self.delete_sibling(x, n + 1, n)
else:
self.delete_merge(x, n, n + 1)
self.delete_predecessor(x.child[n])
/* Delete the successor */
def delete_successor(self, x):
if x.leaf:
return x.keys.pop(0)
if len(x.child[1].keys) >= self.t:
self.delete_sibling(x, 0, 1)
else:
self.delete_merge(x, 0, 1)
self.delete_successor(x.child[0])
/* Delete resolution */
def delete_merge(self, x, i, j):
cnode = x.child[i]
if j > i:
rsnode = x.child[j]
cnode.keys.append(x.keys[i])
for k in range(len(rsnode.keys)):
cnode.keys.append(rsnode.keys[k])
if len(rsnode.child) > 0:
cnode.child.append(rsnode.child[k])
if len(rsnode.child) > 0:
cnode.child.append(rsnode.child.pop())
new = cnode
x.keys.pop(i)
x.child.pop(j)
else:
lsnode = x.child[j]
lsnode.keys.append(x.keys[j])
for i in range(len(cnode.keys)):
lsnode.keys.append(cnode.keys[i])
if len(lsnode.child) > 0:
lsnode.child.append(cnode.child[i])
if len(lsnode.child) > 0:
lsnode.child.append(cnode.child.pop())
new = lsnode
x.keys.pop(j)
x.child.pop(i)
if x == self.root and len(x.keys) == 0:
self.root = new
/* Delete the sibling */
def delete_sibling(self, x, i, j):
cnode = x.child[i]
if i < j:
rsnode = x.child[j]
cnode.keys.append(x.keys[i])
x.keys[i] = rsnode.keys[0]
if len(rsnode.child) > 0:
cnode.child.append(rsnode.child[0])
rsnode.child.pop(0)
rsnode.keys.pop(0)
else:
lsnode = x.child[j]
cnode.keys.insert(0, x.keys[i - 1])
x.keys[i - 1] = lsnode.keys.pop()
if len(lsnode.child) > 0:
cnode.child.insert(0, lsnode.child.pop())
/* Print the tree */
def print_tree(self, x, l=0):
print("Level ", l, " ", len(x.keys), end=":")
for i in x.keys:
print(i, end=" ")
print()
l += 1
if len(x.child) > 0:
for i in x.child:
self.print_tree(i, l)
B = BTree(3)
for i in range(10):
B.insert((i, 2 * i))
B.print_tree(B.root)
B.delete(B.root, (8,))
print("n")
B.print_tree(B.root)Deletion Complexity
Best case Time complexity: Θ(log n)
Average case Space complexity: Θ(n)
Worst case Space complexity: Θ(n)
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