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### Implement special cases here define delete t v let.pdf

• 1. Implement special cases here: (define (delete t v) (let ((path (find t v))) (cond (( empty? t ) ) ((empty? (bst-root t)) ) ((not ( = v (node-value (car path))) ) (( > (node-count (car path)) 1) decrement count) ((equal? (car path) (bst-root t)) ;node were trying to delete (cond ; nest conditiona; ( (let-AND right (car path) empty) ; case 0.1 root <- null ((let-root empty, right empty) ; case 1 root <- left car path ; case 1.1 ((left-empty, right not empty) ; case 1.2 root <- right car path (else (delete-node path)) (else (delete-node path)) ) ) ) ) ) ) ) ) ) Special cases Top-Level Function: Delete Value from BST: (delete t v) The first parameter t is a reference to a bst, and the second parameter v is a value to find and delete. Like the insert function, delete will directly resolve a few special cases. For the general cases, it will call (find t v) to obtain a list of nodes that represent a path from the root to the node to be deleted, and will then call a helper function (delete-node path) to handle the rest of the work. In the case of delete the helper function will further rely on secondary helper functions to resolve cases and subcases. Deletion logically breaks down into cases based on how many child nodes the node to delete has. Each of those cases has subcases. The large number of cases implies the need for a correspondingly larger set of test cases: a test case to trigger each deletion case. Delete Cases Before we look at the detailed code for delete in Racket, lets look at the general approach. As for some of the other functions, it will be convenient to create a top-level function (delete t v) and several helper functions that manage specific detailed cases. There will be some initial special
• 2. cases that should be managed separately, before launching any lower level helper function. Special cases for (delete t v): Tree t is empty: end the function because there is nothing to do. At this point, call (find t v) to create the path from root to node to delete if it is present. Assume the list returned is bound to a symbol named path. The node of interest will be the first node in the list: (car path) By examining node (car path), you can determine if the value to delete is present or not. Value v is not in the tree: end the function because there is nothing to do. Value v is in the tree: o Is the count field > 1? Dont delete the node, simply decrement its count field. o Is the count field = 1? In this case, we must move forward to actually delete the node. Is the node to delete the root of the tree? Does the node have 0 or 1 child nodes? This corresponds to delete cases 0.1, 1.1, and 1.2. Delete the node and update the root of the tree. o If we reach this pointvalue v is in the tree, count field is 1, node is not the root if it has 0 or 1 child nodes then the general delete case applies, call (delete-node path) The remaining delete cases will be covered by the (delete-node path) helper function. Some general considerations for node n (node to delete), node p (parent of n), and node c (child of n). Is node n the left or right child of node p? If node n only has one child, node c, is node c the left child or right child of node n? If node n has two children, we will use an additional procedure to find the in-order predecessor of n, node iop. We will locate node iop, physically delete it, and copy its value and count fields into the original node, node n. This results in a logical but not physical delete of node n. Below is a detailed enumeration of all delete cases based on child count. Subcases 0.1, 1.1, and 1.2 will be handled as special cases by the top-level delete function because these cases cause the root of the tree to change. Since we are maintaining the result of (find t v) in path we can make the following assumptions for all remaining cases: Node to delete: (car path) Parent of node to delete: (car (cdr path)) = (cadr path) Child nodes of node to delete, if they exist: o (node-left(carpath)) o (node-right(carpath)) Path from n to in-order predecessor node: ioppath (only needed if n has two child nodes) In-order predecessor node: (car ioppath) (only needed if node n has two child nodes) Dont confuse path (list of nodes from root of tree to n) with ioppath (list of nodes from n to iop) Case 0: Child Count = 0 Subcase 0.1: parent is null implying node is root (already covered in top-level function) o node being deleted is root of the tree and has 0 child nodes. Set root to null Subcase 0.2: node is left child of parent o set left child of parent to null Subcase 0.3: node is right child of parent o Set right child of parent to null, root doesnt change
• 3. Case 1: Child Count = 1 Subcase 1.1: parent is null and child is left child of node (already covered in top-level) o node is root ,set root to left child Subcase 1.2: parent is null and child is right child of node (already covered in top-level) o Node is root, set root to right child Subcase 1.3: node is left child of parent and child is left child of node o set left child of parent to left child of node Subcase 1.4: node is left child of parent and child is right child of node o set left child of parent to right child of node Subcase 1.5: node is right child of parent and child is left child of node o set right child of parent to left child of node Subcase 1.6: node is right child of parent and child is right child of node o set right child of parent to right child of node The cases and subcases are similar, but still differ in the specifics. Take care to identify the correct case and respond with the correct implementation for that case. In the above cases, note that the root is the only node with no parent (if parent of n is null, then n is the root). Also note the deletion cases that cause a change in the root of the tree (0.1, 1.1, and 1.2) are coded directly in the top- level function. Case 2: Child Count = 2 There is no easy way to delete a node with two child nodes. Instead we substitute the original problem with a problem that is easier to solve: Locate the in-order predecessor of the node to be deleted. Move the value and count fields from the in-order predecessor node to the original node, thereby erasing the previous value and count fields of the original node. This content substitution achieves a logical removal of the original node. Physically remove the in-order predecessor node o Case 2.1: predecessor node is left of parent o Case 2.2: predecessor node is right of paren
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