Java代写数据结构算法 | ACO201 – Data Structures and Algorithms

本次北美CS代写的主要内容是Java相关的BST算法实现

ACO201 – Data Structures and Algorithms Binary Search Tree (BST)

1. Problem Description

You are to develop a simple Binary Search Tree ADT and run it against a test program. Avoid the temptation of finding code online. I am aware of all the available solutions and will be looking closely at your code. You do not need to balance your tree.

  • Basic Level (for a maximum grade of 85%): Develop the BST as a class that uses doubles as keys and Strings as values.
  • Advanced Level (for a maximum grade of 100%): Develop the BST as a generic class that uses any objects as keys and values.

    Remember that a BST is a proper Binary Tree with the following property:
    Let u, v, and w be three nodes such that u is in the left subtree of v and w is in the right subtree of v. We have key(u) < key(v)  key(w)

• Basic Level:
The complete design of the BST class (non-generic) is shown below. You will need to create the inner class Node.

  • ▪  BST(): Construct an empty BST.
  • ▪  void insert(double key, String value): Insert the element (key, value) into the BST into the

    proper position.

  • ▪  String find(double key): Find the first value with the matching key. Return null if the key is not

    found. You may want to define a private helper method find(double, Node) to help with the

    recursive solution.

  • ▪  String delete(double key): Remove the first element with the matching key and return the

    value. Return null if the key is not found.

  • ▪  int size(): Return the number of elements in the BST.

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▪ String toString(): Convert the BST to a hierarchical, indented String, as shown in the output below. Notice that I have also included a String toString(int level) in the Node class to help implement this hierarchy.

• Advanced Level:
The complete design of the generic BST class is shown below. You will need to create the inner class Node. You can assume that keys are Comparable. Refer to the Basic Level description above for a definition of the methods.

With either solution, when your implementation is complete, run it with the test driver BSTTest.java (provided). Review the results to make sure that your BST ADT is working correctly. The expected output is included in the file BSTExpectedOutput.txt and portions are shown below. Your output must include your name.

2. Notes

  • You must use the test driver code found with this assignment: BSTTest.java. You should not modify BSTTest.java except to add your name.
  • Turn in only your source files: BST.java and BSTTest.java. If you have created other classes as part of the solution, you need to turn those in as well.
  • Make sure your class is not in a package (that is, it is in the default package).
  • We normally must be careful when comparing double values. However, for the purposes of this

    assignment, you can assume that d1 == d2 will work for matching double keys.

  • For the generic solution (Advanced level), you can assume that keys are Comparable.
  • You will need to download the file words.txt and put it at the Project level within Eclipse.
  1. Required Main Class

    BSTTest, provided

  2. Required Input

    Not applicable

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5. Required Output

Your output should look like the following. The complete output is too long to include here, so I have removed selected lines (shown as …). The complete output is available in BSTExpectedOutput.txt for comparison purposes. The Large Tree is random so your tree will not match mine.

BST Test Program – Prof. Eckert

        1) Building a Tree
        ========================
          Initially: null
          Insert (3.0, A):
           (k=3.00, v=A)

null

null …

          Insert (4.0, E)
           (k=3.00, v=A)
              (k=2.00, v=C)
                 null
                 null
              (k=5.00, v=B)
                 (k=4.00, v=E)
                    null
                    null
                 (k=6.00, v=D)

null null

        2) Finding Elements
        ========================
          b1.find(2.0): C
           ...
        3) Deleting Elements
        ========================
           ...
          Delete(1.0) Left child, leaf: A
             (k=6.00, v=E)
              (k=3.00, v=H)
                 (k=2.00, v=F)
                    null
                    null
                 (k=5.00, v=C)
                    (k=4.00, v=G)
                       null

null null

              (k=9.00, v=I)
                 (k=7.00, v=J)
                    null
                    (k=8.00, v=B)

null

                       null
                 (k=10.00, v=D)
                    null
                    (k=11.00, v=K)

null null


Delete(5.0) Has left child, internal: C

             (k=6.00, v=E)
              (k=3.00, v=H)
                 (k=2.00, v=F)

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null

            null
         (k=4.00, v=G)

null

            null
      (k=9.00, v=I)
         (k=7.00, v=J)
            null
                       null
                 (k=10.00, v=D)
                    null
                    (k=11.00, v=K)

null null


Delete(6.0) Has two children, internal: E

     (k=7.00, v=J)
      (k=3.00, v=H)
                 (k=2.00, v=F)
                    null
            null
         (k=4.00, v=G)

null

            null
      (k=9.00, v=I)
                    null
                    (k=11.00, v=K)

null null

4) A Large Tree
========================
  First 10: Size is now 10
   ...

Remaining Elements: Size is now 100000

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