Binary Search Trees Rem Collier Room A1.02 School of Computer Science and Informatics University College Dublin, Ireland

Binary Search Tree A Binary Search Tree is a Binary Tree that satisfies the following properties: Each internal node holds a (unique) value. A total-order relation (~) is defined on those values. Reflexive: k ~ k Antisymmetric: if k1 ~ k2 and k2 ~ k1 then k1 = k2 Transitive: if k1 ~ k2 and k2 ~ k3 then k1 ~ k3 All the values (ki) in the left sub-tree of a node with value k satisfy the relation ki ~ k. All the values (kj) in the right sub-tree of a node with value k satisfy the relation k ~ kj. Useful Feature: An in-order traversal of a binary search trees visits the values in the order specified by the total-order relation.

Binary Search Tree Example Consider a Binary Search Tree for integer values, using the less than (≤) total order relation. The state of a Binary Search Tree depends on the order in which items are added. 6 9 2 4 1 8

Uses of BSTs Binary Search Trees are an implementation strategy for other ADTs. Support three basic operations: insert(e): Add a new node containing e, if one does not already exist find(e): Find the node containing e remove(e): Remove the node containing e Can easily be adapted to store entries and support retrieval by key. Do you know of any other ADTs that have similar operations?

Search To search for a value k, we trace a downward path starting at the root. The next node visited depends on the outcome of the comparison of k with the value of the current node If we reach a leaf, the key is not found and we return null Example: find(4) Algorithm find(k, v) if T.isExternal(v) return null if k < v.element() return find(k, T.leftChild(v)) else if k = v.element() return v else return find(k, T.rightChild(v)) < 6 2 > 9 1 4 = 8

Insertion To perform operation insert(k), we search for value k Assume k is not already in the tree, and let let w be the leaf reached by the search We insert k at node w and expand w into an internal node Example: insert(5) < 6 2 9 > 1 4 8 > w 6 2 9 1 4 8 w 5

Deletion To perform operation remove(k), we search for k. Assume k is in the tree, and let let v be the node storing k If node v has one child, then we can remove it using the existing remove(v) operation. Example: remove(4) 6 < 2 9 > v 1 4 8 w 5 6 2 9 1 5 8

Deletion (cont.) What about the case where the k to be removed is stored at a node v that has 2 children? we find the internal node w that follows v in an inorder traversal we copy w.element() into node v we remove node w and its left child z (which must be a leaf) Example: remove(3) 1 v 3 2 8 6 9 w 5 z 1 v 5 2 8 6 9

Problems Create the following BST: insert(15), insert(25), insert(5), insert(9), remove(15) insert(37), insert(26), insert(35), insert(28), remove(26) insert(40), insert(1), insert(15), insert(26), insert(6), remove(5) insert(H), insert(A), insert(P), insert(Y), insert(O), remove(H) insert(B), insert(C), insert(F), insert(E), remove(A) insert(H), insert(A)

Performance Consider a binary tree containing n items that is of height h the space used is O(n) methods find(), insert() and remove() take O(h) time The height h is O(n) in the worst case and O(log n) in the best case! The best case arises when the tree is balanced!