Algorithm of Binary Tree - EPI


Find the first occurrence of k in a BST

SearchBSTForFirstOccurrenceRecursive.java

  public static BinaryTree<Integer> findFirstEqualK(

      BinaryTree<Integer> T, Integer k) {

    if (T == null) {

      return null; // No match.

    } else if (T.getData().compareTo(k) == 0) {

      // Recursively searches the left subtree for first one == k.

      BinaryTree<Integer> n = findFirstEqualK(T.getLeft(), k);

      return n != null ? n : T;

    }

    // Searches left or right tree according to T.getData() and k.

    return findFirstEqualK(

        T.getData().compareTo(k) < 0 ? T.getRight() : T.getLeft(), k);

  }

SearchBSTForFirstOccurrenceIterative.java

  public static BinaryTree<Integer> findFirstEqualK(
      BinaryTree<Integer> T, Integer k) {
    BinaryTree<Integer> first = null;
    BinaryTree<Integer> curr = T;
    while (curr != null) {
      if (curr.getData().compareTo(k) < 0) {
        curr = curr.getRight();
      } else if (curr.getData().compareTo(k) > 0) {
        curr = curr.getLeft();
      } else { // curr.getData().compareTo(k) == 0
        // Searches for the leftmost in the left subtree.
        first = curr;
        curr = curr.getLeft();
      }
    }
    return first;
  }

Find the first key larger than k in a BST

SearchBSTFirstLargerK.java

  public static BinaryTree<Integer>
  findFirstLargerKWithKExist(BinaryTree<Integer> r, Integer k) {
    boolean foundK = false;
    BinaryTree<Integer> curr = r;
    BinaryTree<Integer> first = null;

    while (curr != null) {
      if (curr.getData().compareTo(k) == 0) {
        foundK = true;
        curr = curr.getRight();
      } else if (curr.getData().compareTo(k) > 0) {
        first = curr;
        curr = curr.getLeft();
      } else { // curr.getData().compareTo(k) < 0
        curr = curr.getRight();
      }
    }
    return foundK ? first : null;
  }

Find the k largest elements in a BST  FindKLargestBST.java

  public static List<Integer> findKLargestInBST(
      BinaryTree<Integer> root, int k) {
    List<Integer> kElements = new ArrayList<>();
    findKLargestInBSTHelper(root, k, kElements);
    return kElements;
  }

  private static void findKLargestInBSTHelper(
      BinaryTree<Integer> r, int k, List<Integer> kElements) {
    // Performs reverse inorder traversal.
    if (r != null && kElements.size() < k) {
      findKLargestInBSTHelper(r.getRight(), k, kElements);
      if (kElements.size() < k) {
        kElements.add(r.getData());
        findKLargestInBSTHelper(r.getLeft(), k, kElements);
      }
    }
  }

Build a BST from a sorted array with minimum height BuildBSTFromSortedSortedArray.java

  private static BinaryTree<Integer> buildBSTFromSortedArrayHelper(
      List<Integer> a, int start, int end) {
    if (start < end) {
      int mid = start + ((end - start) / 2);
      return new BinaryTree<>(a.get(mid),
          buildBSTFromSortedArrayHelper(a, start, mid),
          buildBSTFromSortedArrayHelper(a, mid + 1, end));
    }
    return null;
  }

Update a BST  UpdateBST.java


Compute the average of the top three scores AverageTop3Scores.java

To save memory, we can save at most 3 scores for each student: when add scores, if there are already 3 scores, compare current score with least score in the map, if current score is bigger, remove the least one and add current score, otherwise ignore it.
  private static class UniqueInteger implements Comparable<UniqueInteger> {
    public Integer value;
    public Long sequence;

    public UniqueInteger(Integer value, Long sequence) {
      this.value = value;
      this.sequence = sequence;
    }
    public int compareTo(UniqueInteger o) {
      int result = value.compareTo(o.value);
      if (result == 0) {
        result = sequence.compareTo(o.sequence);
      }
      return result;
    }
  }
  public static String findStudentWithTopThreeAverageScores(InputStream ifs) {
    Map<String, TreeSet<UniqueInteger>> studentScores = new HashMap<>();
    try {
      long sequence = 0;
      ObjectInputStream ois = new ObjectInputStream(ifs);
      while (true) {
        String name = ois.readUTF();
        int score = ois.readInt();
        TreeSet<UniqueInteger> scores = studentScores.get(name);
        if (scores == null) {
          scores = new TreeSet<>();
        }
        scores.add(new UniqueInteger(score, sequence++));
        studentScores.put(name, scores);
      }
    } catch (IOException e) {
    }

    String topStudent = "no such student";
    int currentTopThreeScoresSum = 0;
    for (Map.Entry<String, TreeSet<UniqueInteger>> scores : studentScores
        .entrySet()) {
      if (scores.getValue().size() >= 3) {
        int currentScoresSum = getTopThreeScoresSum(scores.getValue());
        if (currentScoresSum > currentTopThreeScoresSum) {
          currentTopThreeScoresSum = currentScoresSum;
          topStudent = scores.getKey();
        }
      }
    }
    return topStudent;
  }

  // Returns the sum of top three scores.
  private static int getTopThreeScoresSum(TreeSet<UniqueInteger> scores) {
    Iterator<UniqueInteger> it = scores.iterator();
    int result = 0;
    for (int i = 0; i < 3 && it.hasNext(); i++) {
      result += it.next().value;
    }
    return result;
  }

Test if a binary tree is a min-first BST SearchMinFirstBST.java

  public static boolean searchMinFirstBST(
      BinaryTree<Integer> T, Integer k) {
    if (T == null || T.getData().compareTo(k) > 0) {
      return false;
    } else if (T.getData().compareTo(k) == 0) {
      return true;
    }

    // Search the right subtree if the smallest key in the right subtree is
    // greater than or equal to k.
    if (T.getRight() != null && k.compareTo(T.getRight().getData()) >= 0) {
      return searchMinFirstBST(T.getRight(), k);
    }
    return searchMinFirstBST(T.getLeft(), k);
  }

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