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Problem 1 Assume the priority queue also must support the update DELETE(A,i), where i gives you the location in the data strucure where element i is located. In a binary heap, that would be A [ i ]. Write pseudocode to achieve DELETE(A,i) in O (log n ) time for binary heaps, where n is the size of the heap. Solution: 1, Delete a node from the array (this creates a vacant node and the tree is not complete) 2. Replace the deletion node with the fartest right node on the lowest level of the Binary Tree (To complete the tree) 3. Heapify (fix the heap): Algorithm for Max Heapify Procedure: MaxHeapify(A, n) 1. l left(n) 2. r right(n) 3. if l heap-size[A] and A[l] > A[n] 4. then largest l 5. else largest i 6. if r heap-size[A] and A[r] > A[largest] 7. then largest r 8. if largest n 9. then exchange A[n] A[largest] 10. MaxHeapify(A, largest) Running time( delete(n) ) = Running time heap delete in heap with n node = height of a heap with n nodes height of the heap containing n nodes =log(n + 1) Running time of delete into a heap of n nodes = O(log(n)) Complexity of deletion: O(logN) in the worst case.
Problem 2 Using only the definition of a binary search tree, show that if a node in a binary search tree has two children, then its successor has no left child and then its predecessor has no right child.

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• Fall '16

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