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CS 61B: Lecture 26
Wednesday, October 27, 2010
Today’s reading: Goodrich & Tamassia, Section 10.1.
Representing Binary Trees
-------------------------
Recall that a binary tree is a rooted tree wherein no node has more than
two children. Additionally, every child is either a _left_child_ or
a _right_child_ of its parent, even if it is its parent’s only child.
In the most popular binary tree representation, each tree node has three
references to neighboring tree nodes: a "parent" reference, and "left" and
"right" references for the two children. (For some algorithms, the "parent"
references are unnecessary.) Each node also has an "item" reference.
public class BinaryTreeNode { | public class BinaryTree {
Object item; | BinaryTreeNode root;
BinaryTreeNode parent; | int size;
BinaryTreeNode left; | }
BinaryTreeNode right; |
public void inorder() {
if (left != null) {
left.inorder();
}
this.visit();
if (right != null) {
right.inorder();
}
}
}
================================================
+ BINARY TREE | ------------------- +
=============== |--- --- | +
+ ||.|root size|7| | +
+ |-+- --- | +
+ --|---------------- +
+ v BinaryTree object +
+ ----- +
+ | * | +
+ ----- ------------ +
+ Root node => |add| | parent | +
+ ----- ------------ +
+ |.|.| | item | +
+ /---\ ------------ +
+ / ^^ \ |left|right| +
+ v / \ v ------------ +
+ ---/- -\--- structure of +
+ | . | | . | BinaryTreeNodes +
+ ----- ----- +
+ |sub| |div| +
+ ----- ----- +
+ >|.|.| |.|.|< +
+ / /--|- -|--\ \ +
+ / / ^| |^ \ \ +
+ / v |v v| v \ +
+ --+-- --+-- --+-- --+-- +
+ | . | | . | | . | | . | +
+ ----- ----- ----- ----- +
+ | 6 | | 5 | | 9 | | 3 | +
+ ----- ----- ----- ----- +
+ |*|*| |*|*| |*|*| |*|*| +
+ ----- ----- ----- ----- +
================================================
BINARY SEARCH TREES
===================
An _ordered_dictionary_ is a dictionary in which the keys have a total order,
just like in a heap. You can insert, find, and remove entries, just as with a
hash table. But unlike a hash table, you can quickly find the entry with
minimum or maximum key, or the entry nearest another entry in the total order.
An ordered dictionary does anything a dictionary or binary heap can do and
more, albeit more slowly.
The simplest implementation of an ordered dictionary is a binary search tree,
wherein entries are maintained in a (somewhat) sorted order.
18 The _left_subtree_ of a node is the subtree rooted at the
/ \ node’s left child; _right_subtree_ is defined similarly.
12 25 A binary search tree satisfies the _binary_search_tree_
/ \ / \ _invariant_: for any node X, every key in the left subtree
4 15 25 30 of X is less than or equal to X’s key, and every key in the
/ / \ / right subtree of X is greater than or equal to X’s key. You
1 13 17 28 can verify this in the search tree at left: for instance,
\ \ \ the root is 18, its left subtree (rooted at 12) contains
3 14 29 numbers from 1 to 17, and its right subtree (rooted at 25)
contains numbers from 25 to 30.
When a node has only one child, that child is either a left child or a right
child, depending on whether its key is smaller or larger than its parent’s key.
(A key equal to the parent’s key can go into either subtree.)
An inorder traversal of a binary search tree visits the nodes in sorted order.
In this sense, a search tree maintains a sorted list of entries. However,
operations on a search tree are usually more efficient than the same operations
on a sorted linked list.
Let’s replace the "Object item;" declaration in each node with "Entry entry;"
where each Entry object stores a key and an associated value. The keys
implement the Comparable interface, and the key.compareTo() method induces a
total order on the keys (e.g. alphabetical order).
[1] Entry find(Object k);
public Entry find(Object k) {
BinaryTreeNode node = root; // Start at the root.
while (node != null) {
int comp = ((Comparable) k).compareTo(node.entry.key());
if (comp < 0) { // Repeatedly compare search
node = node.left; // key k with current node; if
} else if (comp > 0) { // k is smaller, go to the left
node = node.right; // child; if k is larger, go to
} else { /* The keys are equal */ // the right child. Stop when
return node.entry; // we find a match (success;
} // return the entry) or reach
} // a null pointer (failure;
return null; // return null).
}
This method only finds exact matches. What if we want to find the smallest key
greater than or equal to k, or the largest key less than or equal to k?
Fortunately, when searching downward through the tree for a key k that is not
in the tree, we are certain to encounter both
- the node containing the smallest key greater than k (if any key is greater)
- the node containing the largest key less than k (if any key is less).
See Footnote 1 for an explanation why.
