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Show Source |    | About   «  17.3. Binary Trees Part 1   ::   Contents   ::   17.5. Binary Trees Part 3  »

17.4. Binary Trees Part 2

17.4.1. Binary Trees Part 2

17.4.1.1. Full and Complete Binary Trees

Full binary tree: Each node is either a leaf or internal node with exactly two non-empty children.

Complete binary tree: If the height of the tree is \(d\), then all leaves except possibly level \(d\) are completely full. The bottom level has all nodes to the left side.

17.4.1.2. Full Binary Tree Theorem (1)

Theorem: The number of leaves in a non-empty full binary tree is one more than the number of internal nodes.

Proof (by Mathematical Induction):

Base case: A full binary tree with 1 internal node must have two leaf nodes.

Induction Hypothesis: Assume any full binary tree T containing \(n-1\) internal nodes has \(n\) leaves.

17.4.1.3. Full Binary Tree Theorem (2)

Induction Step: Given tree T with \(n\) internal nodes, pick internal node \(I\) with two leaf children. Remove \(I\)’s children, call resulting tree T’.

By induction hypothesis, T’ is a full binary tree with \(n\) leaves.

Restore \(I\)’s two children. The number of internal nodes has now gone up by 1 to reach \(n\). The number of leaves has also gone up by 1.

17.4.1.4. Full Binary Tree Corollary

Theorem: The number of null pointers in a non-empty tree is one more than the number of nodes in the tree.

Proof: Replace all null pointers with a pointer to an empty leaf node. This is a full binary tree.

17.4.1.5. Dictionary

/** The Dictionary abstract class. */
public interface Dictionary {

  /** Reinitialize dictionary */
  public void clear();

  /** Insert a record
      @param key The key for the record being inserted.
      @param elem The record being inserted. */
  public void insert(Comparable key, Object elem);

  /** Remove and return a record.
      @param key The key of the record to be removed.
      @return A maching record. If multiple records match
      "k", remove an arbitrary one. Return null if no record
      with key "k" exists. */
  public Object remove(Comparable key);

  /** Remove and return an arbitrary record from dictionary.
      @return the record removed, or null if none exists. */
  public Object removeAny();

  /** @return A record matching "k" (null if none exists).
      If multiple records match, return an arbitrary one.
      @param key The key of the record to find */
  public Object find(Comparable key);

  /** @return The number of records in the dictionary. */
  public int size();
}
/** The Dictionary abstract class. */
public interface Dictionary<K, E> {

  /** Reinitialize dictionary */
  public void clear();

  /** Insert a record
      @param k The key for the record being inserted.
      @param e The record being inserted. */
  public void insert(K key, E elem);

  /** Remove and return a record.
      @param k The key of the record to be removed.
      @return A maching record. If multiple records match
      "k", remove an arbitrary one. Return null if no record
      with key "k" exists. */
  public E remove(K key);

  /** Remove and return an arbitrary record from dictionary.
      @return the record removed, or null if none exists. */
  public E removeAny();

  /** @return A record matching "k" (null if none exists).
      If multiple records match, return an arbitrary one.
      @param k The key of the record to find */
  public E find(K key);

  /** @return The number of records in the dictionary. */
  public int size();
}

17.4.1.6. .

.

17.4.1.7. Dictionary (2)

  • How can we implement a dictionary?

    • We know about array-based lists and linked lists.

    • They might be sorted or unsorted.

    • What are the pros and cons?

17.4.1.8. Binary Search Trees

17.4.1.9. BST as a Dictionary (1)

17.4.1.10. BST as a Dictionary (2)

17.4.1.11. BST findhelp

Settings

Proficient Saving... Error Saving
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17.4.1.12. BST inserthelp

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17.4.1.13. BST deletemax

Settings

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17.4.1.14. BST removehelp

Settings

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Server Error
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17.4.1.15. .

.

17.4.1.16. BST Analysis

Find: \(O(d)\)

Insert: \(O(d)\)

Delete: \(O(d)\)

\(d =\) depth of the tree

\(d\) is \(O(\log n)\) if the tree is balanced.

What is the worst case cost? When?

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