Introduction to B-Trees A B-tree is a tree data structure that keeps data sorted and allows searches, insertions, and deletions in logarithmic amortized time. Unlike self-balancing binary search trees, it is optimized for systems that read and write large blocks of data. It is most commonly used in database and file systems. The B-Tree Rules Important properties of a B-tree: B-tree nodes have many more than two children. A B-tree node may contain more than just a single element. The set formulation of the B-tree rules: Every B-tree depends on a positive constant integer called MINIMUM, which is used to determine how many elements are held in a single node. Rule 1: The root can have as few as one element (or even no elements if it also has no children); every other node has at least MINIMUM elements. Rule 2: The maximum number of elements in a node is twice the value of MINIMUM. Rule 3: The elements of each B-tree node are stored in a partially filled array, sorted from the smallest element (at index 0) to the largest element (at the final used position of the array). Rule 4: The number of subtrees below a nonleaf node is always one more than the number of elements in the node. Subtree 0, subtree 1, ... Rule 5: For any nonleaf node: An element at index i is greater than all the elements in subtree number i of the node, and An element at index i is less than all the elements in subtree number i + 1 of the node. Rule 6: Every leaf in a B-tree has the same depth. Thus it ensures that a B-tree avoids  the problem of a unbalanced tree. Searching for a Target in a Set The psuedocode: Make a local variable, i, equal to the first index such that data[i] >= target. If there is no such index, then set i equal to dataCount, indicating that none of the elements is greater than or equal to the target. if (we found the target at data[i]) return true; else if (the root has no children) return false;else return subset[i].contains(target); Adding an Element to a B-Tree It is easier to add a new element to a B-tree if we relax one of the B-tree rules. Loose addition allows the root node of the B-tree to have MAXIMUM + 1 elements. For example, suppose we want to add 18 to the tree: The above result is an illegal B-tree. Our plan is to perform a loose addition first, and then fix the root's problem. The Loose Addition Operation for a B-Tree: private void looseAdd(int element) {     1. i = firstGE(element) // find the first index such that data[i] >= element     2. if (we found the new element at data[i]) return; // since there's already a copy in the set     3. else if (the root has no children)            Add the new element to the root at data[i]. (shift array)     4. else {            subset[i].looseAdd(element);            if the root of subset[i] now has an excess element, then fix that problem before returning.        }} private void fixExcess(int i) // precondition: (i < childCount) and the entire B-tree is valid except that subset[i] has MAXIMUM + 1 elements. // postcondition: the tree is rearranged to satisfy the loose addition rule Fixing a Child with an Excess Element: To fix a child with MAXIMIM + 1 elements, the child node is split into two nodes that each contain MINIMUM elements. This leaves one extra element, which is passed up to the parent. It is always the middle element of the split node that moves upward. The parent of the split node gains one additional child and one additional element. The children of the split node have been equally distributed between the two smaller nodes.  Fixing the Root with an Excess Element: Create a new root. fixExcess(0). Removing an Element from a B-Tree Loose removal rule: Loose removal allows to leave a root that has one element too few. public boolean remove(int target) {     answer = looseRemove(target);     if ((dataCount == 0) && (childCount == 1))         Fix the root of the entire tree so that it no longer has zero elements;     return answer; } private boolean looseRemove(int target) { 1. i = firstGE(target) 2. Deal with one of these four possibilities:    2a. if (root has no children and target not found) return false.    2b. if( root has no children but target found) {            remove the target            return true        }    2c. if (root has children and target not found) {            answer = subset[i].looseRemove(target)            if (subset[i].dataCount < MINIMUM)                fixShortage(i)            return true        }    2d. if (root has children and target found) {            data[i] = subset[i].removeBiggest()            if (subset[i].dataCount < MINIMUM)                fixShortage(i)            return true       } } private void fixShortage(int i) // Precondition: (i < childCount) and the entire B-tree is valid except that subset[i] has MINIMUM - 1 elements. // Postcondition: problem fixed based on the looseRemoval rule. private int removeBiggest() // Precondition: (dataCount > 0) and this entire B-tree is valid // Postcondition: the largest element in this set has been removed and returned. The entire B-tree is still valid based on the looseRemoval rule. Fixing Shortage in a Child: When fixShortage(i) is activated, we know that subset[i] has MINIMUM - 1 elements. There are four cases that we need to consider: Case 1: Transfer an extra element from subset[i-1]. Suppose subset[i-1] has more than the MINIMUM number of elements. Transfer data[i-1] down to the front of subset[i].data. Transfer the final element of subset[i-1].data up to replace data[i-1]. If subset[i-1] has children, transfer the final child of subset[i-1] over to the front of subset[i].  Case 2: Transfer an extra element from subset[i+1]. Suppose subset[i+1] has more than the MINIMUM number of elements. Case 3: Combine subset[i] with subset[i-1]. Suppose subset[i-1] has only MINIMUM elements. Transfer data[i-1] down to the end of subset[i-1].data. Transfer all the elements and children from subset[i] to the end of subset[i-1]. Disconnect the node subset[i] from the B-tree by shifting subset[i+1], subset[i+2] and so on leftward.  Case 4: Combine subset[i] with subset[i+1]. Suppose subset[i+1] has only MINIMUM elements. We may need to continue activating fixShortage() until the B-tree rules are satisfied. Removing the Biggest Element from a B-Tree: private int removeBiggest() {     if (root has no children)         remove and return the last element     else {         answer = subset[childCount-1].removeBiggest()         if (subset[childCount-1].dataCount < MINIMUM)             fixShortage(childCount-1)         return answer     } } A more concrete example for node deletion:  