First: Implement the class TreapADT including an search, insert, and delete methods.
Then split the tree by adding a SplitTree method.
✔ The tree splitting problem is this:
✗ Given a tree and a key value K not in the tree, create two trees: One with keys less than K, and one with keys greater than K
✔ This is easy to solve with a treap, once the insert operation has been implemented:
✗ Insert (K, INFINITY) in the treap
✗ Since this has a higher priority than any node in the heap, it will become the root of the treap after insertion
✗ Because of the BST ordering property, the left subtree of the root will be a treap with keys less than K, and the right subtree of the root will be a treap with keys greater than K. These subtrees then are the desired result of the split
✔ Since insert can be done in time O(H) where H is the height of the treap, splitting can also be done in time O(H)
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split a bst in to 2 trees
First: Implement the class TreapADT including an search, insert, and delete methods.
Then split the tree by adding a SplitTree method.
✔ The tree splitting problem is this:
✗ Given a tree and a key value K not in the tree, create two trees: One with keys less than K, and one with keys greater than K
✔ This is easy to solve with a treap, once the insert operation has been implemented:
✗ Insert (K, INFINITY) in the treap
✗ Since this has a higher priority than any node in the heap, it will become the root of the treap after insertion
✗ Because of the BST ordering property, the left subtree of the root will be a treap with keys less than K, and the right subtree of the root will be a treap with keys greater than K. These subtrees then are the desired result of the split
✔ Since insert can be done in time O(H) where H is the height of the treap, splitting can also be done in time O(H)
Place your Order Now
First: Implement the class TreapADT including an search, insert, and delete methods.
Then split the tree by adding a SplitTree method.
✔ The tree splitting problem is this:
✗ Given a tree and a key value K not in the tree, create two trees: One with keys less than K, and one with keys greater than K
✔ This is easy to solve with a treap, once the insert operation has been implemented:
✗ Insert (K, INFINITY) in the treap
✗ Since this has a higher priority than any node in the heap, it will become the root of the treap after insertion
✗ Because of the BST ordering property, the left subtree of the root will be a treap with keys less than K, and the right subtree of the root will be a treap with keys greater than K. These subtrees then are the desired result of the split
✔ Since insert can be done in time O(H) where H is the height of the treap, splitting can also be done in time O(H)
– Essay Answers | www.essayanswers.org