1 Trees - Part II © Dave Bockus

2 Building a Binary Search Tree h i b c e d f m k a Input: i h b m e c f a d k

3 Code for BST Insertion Using Recursion BinaryNode Insert(Comparable x, BinaryNode t) { if (t == null) Null ptr, Create new Node. t = new BinaryNode(X, null, null); else if (x.compareTo(t.element < 0)If element is < then descend left t.left = Insert(x, t.left); else if (x.compareTo(t.element > 0)If element is > then descend right t.right = Insert(x, t.right); else ; Equal i.e. Duplicate - do nothing return t; }

4 BST Insertion Using Iteration - page 308 Lafore public void insert(int id, double dd) { Node newNode = new Node(); // make new node newNode.iData = id; // insert data newNode.dData = dd; if(root==null) // no node in root root = newNode; else { // root occupied Node current = root; // start at root Node parent; while(true) { // (exits internally) parent = current; if(id < current.iData){ // go left? current = current.leftChild; if(current == null) { // if end of the line parent.leftChild = newNode; // insert on left return; } } // end if go left else { // or go right? current = current.rightChild; if(current == null){ // if end of the line parent.rightChild = newNode; // insert on right return; } } // end else go right } // end while } // end else not root } // end insert()

5 Modified BST Insertion Using Iteration public Node insert(int id) //Data is modified outside of {// insert via returned ptr Node newNode = new Node(); // make new node newNode.iData = id; // insert key data if(root==null) // no node in root root = newNode; return root; else { // root occupied Node current = root; // start at root Node parent; while(true) { // (exits internally) parent = current; if (id == current.iData) return current; if(id < current.iData){ // go left? current = current.leftChild; if(current == null) { // if end of the line parent.leftChild = newNode; // insert on left return parent.leftChild; } } // end if go left else { // or go right? current = current.rightChild; if(current == null){ // if end of the line parent.rightChild = newNode; // insert on right return parent.rightChild; } } // end else go right } // end while } // end else not root } // end insert()

6 Building an AVL Tree J M B G E Input: M J B F E G Single Right J B M F Double Left B E F Double Right F JE GMB

7 Deletions in a Binary Tree 3 Cases - if we delete node pointed to by ptr –Leaf Node Just Delete it –Node with 1 null sub-tree. Ptr’s Parent points to non-null sub-tree. –Node with 2 non-null sub-trees. Find successor Copy successor into node pointed to by ptr. Delete successor - simple case i.e. null left sub-tree

8 Deletion Case 1 - A leaf node h i b c e d f m k a Leaf nodes can be deleted without any problems

9 h i b c e d f m k a Deletion Case non-null sub-tree Parent points to non-null subtree ptr Node pointed to by ptr is deleted

10 h i b c e d f m k a Deletion Case non-null sub-trees c ptr Find the successor Replace data of ptr with data of qtr Deleting node qtr is a simple case qtr c

11 Code to Delete from a BST BinaryNode remove(Comparable x, BinaryNode t) { if (t == null) Null ptr, node not found return t; if (x.compareTo(t.element) < 0)Descend left branch t.left = remove(x, t.left); else if(x.compareTo(t.element) > 0)Descend right branch t.right = remove(x, t.right); else if (t.left != null && t.right != null) { t.element = successor(t).element; Copy successor into current t.right = remove(t.element, t.right); node and recursively remove } the successor. else if (t.left != null) Simple case: found x pointed to by t = t.left;t, so point around it to the non- elsenull sub-tree. If t is a leaf then t = t.right;t.left is null so we point around to return t;t.right which is null also, this }null is returned.