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pass # nothing to remove else if x.key < root.item.key: root.left <- TREE-DELETE(root.left, x) else if x.key > root.item.key: root.right <- TREE-DELETE (root.right, x) # х.key if root.left is NIL: root <- root.right else: == root.item.key so Remove root.item. # NIL if both children missing else if root.right is NIL: root <- root.left else: # Root has two children: remove element with smallest key in right subtree and # move it to root. Assumption: DELETE-MIN returns two values: element removed # from right subtree and root of resulting subtree (in case root.right changes). root.item, root.right <- DELETE-MIN(root.right) return root DELETE-MIN(root): # Remove element with smallest key in root’s subtree; # return element removed and root of resulting subtree. if root.left is NIL: # Root stores item with smallest key; replace with right child. return root.item, root.right else: # Left subtree not empty: root not the smallest. item, root.left <- DELETE-MIN(root.left) return item, root

pass # nothing to remove else if x.key < root.item.key: root.left <- TREE-DELETE(root.left, x) else if x.key > root.item.key: root.right <- TREE-DELETE (root.right, x) # х.key if root.left is NIL: root <- root.right else: == root.item.key so Remove root.item. # NIL if both children missing else if root.right is NIL: root <- root.left else: # Root has two children: remove element with smallest key in right subtree and # move it to root. Assumption: DELETE-MIN returns two values: element removed # from right subtree and root of resulting subtree (in case root.right changes). root.item, root.right <- DELETE-MIN(root.right) return root DELETE-MIN(root): # Remove element with smallest key in root’s subtree; # return element removed and root of resulting subtree. if root.left is NIL: # Root stores item with smallest key; replace with right child. return root.item, root.right else: # Left subtree not empty: root not the smallest. item, root.left <- DELETE-MIN(root.left) return item, root

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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TREE-DELETE(root, x): should not happen! # x.key not in S # nothing to remove if root is NIL: pass else if x.key < root.item.key: root.left <- TREE-DELETE(root.left, x) else if x.key > root.item.key: root.right <- TREE-DELETE (root.right, x) # x.key if root.left is NIL: root <- root.right else if root.right is NIL: else: root.item.key so Remove root.item. == # NIL if both children missing root <- root.left else: # Root has two children: remove element with smallest key in right subtree and # move it to root. Assumption: DELETE-MIN returns two values: element removed # from right subtree and root of resulting subtree (in case root.right changes). root.item, root.right <- DELETE-MIN(root.right) return root DELETE-MIN (root): # Remove element with smallest key in root's subtree; # return element removed and root of resulting subtree. if root.left is NIL: # Root stores item with smallest key; replace with right child. return root.item, root.right else: # Left subtree not empty: root not the smallest. item, root.left <- DELETE-MIN(root.left) return item, root
Transcribed Image Text:TREE-DELETE(root, x): should not happen! # x.key not in S # nothing to remove if root is NIL: pass else if x.key < root.item.key: root.left <- TREE-DELETE(root.left, x) else if x.key > root.item.key: root.right <- TREE-DELETE (root.right, x) # x.key if root.left is NIL: root <- root.right else if root.right is NIL: else: root.item.key so Remove root.item. == # NIL if both children missing root <- root.left else: # Root has two children: remove element with smallest key in right subtree and # move it to root. Assumption: DELETE-MIN returns two values: element removed # from right subtree and root of resulting subtree (in case root.right changes). root.item, root.right <- DELETE-MIN(root.right) return root DELETE-MIN (root): # Remove element with smallest key in root's subtree; # return element removed and root of resulting subtree. if root.left is NIL: # Root stores item with smallest key; replace with right child. return root.item, root.right else: # Left subtree not empty: root not the smallest. item, root.left <- DELETE-MIN(root.left) return item, root
2. BST Deletion Average Case (b) We are interested in the total number of calls to TREE-DELETE and DELETE-MIN that will be performed to delete nodes from this tree. The code for these functions is distributed with this test in the file bst-deletions.pdf. It is also the same code that was taught in week 3 lectures. How many calls would be made to these two functions (in total) if we deleted 2? (c) How many calls will be made to these functions if we deleted 5 (from the original tree that still holds the 2)? (d) Assume now that we will randomly select one node from the tree for deletion. Assume that the probability of selecting the 7 is and the probability of selecting the 8 is . The other nodes in the tree are equally likely to be selected for deletion and the probability of calling delete on a node that is not in the tree is 0. Calculate the average-case number of total calls to TREE-DELETE and DELETE-MIN. Show your work.
Transcribed Image Text:2. BST Deletion Average Case (b) We are interested in the total number of calls to TREE-DELETE and DELETE-MIN that will be performed to delete nodes from this tree. The code for these functions is distributed with this test in the file bst-deletions.pdf. It is also the same code that was taught in week 3 lectures. How many calls would be made to these two functions (in total) if we deleted 2? (c) How many calls will be made to these functions if we deleted 5 (from the original tree that still holds the 2)? (d) Assume now that we will randomly select one node from the tree for deletion. Assume that the probability of selecting the 7 is and the probability of selecting the 8 is . The other nodes in the tree are equally likely to be selected for deletion and the probability of calling delete on a node that is not in the tree is 0. Calculate the average-case number of total calls to TREE-DELETE and DELETE-MIN. Show your work.
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