We have already seen stacks and queues, two abstract data types. We've seen that it's possible to implement these abstract types using various concrete data structures. For example, we can implement a stack or queue either using an array or a linked list.
Let's now introduce another abstract data type. A set provides the following operations:
A set cannot contain the same value twice: every value is either present in the set or it is not.
This type should be familiar, because it has a default implementation in Python that we have seen and used.
In this algorithms course, we'll study various ways to implement sets. We'd like to understand how we could build efficient sets in Python if they were not already provided in the standard library.
A binary tree consists of a set of nodes. Each node contains a single value and may have 0, 1, or 2 children.
Here is a picture of a binary tree of integers. (Note that this is not a binary search tree, which is a special kind of binary tree that we will discuss later.)
In this tree, node 10 is the root node. 14 is the parent of 12 and 6. 12 is the left child of 14, and 6 is the right child of 14. 14 is an ancestor of 22, and 22 is a descendant of 14.
A node may have 0, 1, or 2 children. In this tree, node 15 has a right child but no left child.
The subtree rooted at 14 is the left subtree of node 10. The subtree rooted at 1 is the right subtree of node 10.
The nodes 12, 5, 22, 4, and 3 are leaves: they have no children. Nodes 10, 14, 1, 6, and 15 are internal nodes, which are nodes that have at least one child.
The depth of a node is its distance from the root. The root has depth 0. In this tree, node 15 has depth 2 and node 4 has depth 3. The height of a tree is the greatest depth of any node. This tree has height 3.
The tree with no nodes is called the empty tree.
Note that a binary tree may be asymmetric: the right side might not look at all like the left. In fact a binary tree can have any structure at all, as long as each node has 0, 1, or 2 children.
A binary tree is complete iff every internal node has 2 children and all leaves have the same height. Here is a complete binary tree of height 3:
A complete binary tree of height h has 2h leaves, and has 20 + 21 + … + 2h-1 = 2h – 1 interior nodes. So it has a total of 2h + 2h – 1 = 2h + 1 – 1 nodes. In this tree there are 23 = 8 leaves and 23 – 1 = 7 interior nodes, a total of 24 – 1 = 15 nodes.
Conversely if a complete binary tree has N nodes, then N = 2h + 1 – 1, where h is the height of the tree. And so h = log2(N + 1) – 1 ≈ log2(N) – 1 = O(log N).
We can represent a binary tree in Python using an object for each node, similarly to how we represent linked lists. Here is a Node class we can use for binary trees:
class Node:
def __init__(self, val, left, right):
self.val = val
self.left = left # left child, or None if absent
self.right = right # right child, or None if absent
We will generally refer to a tree using a pointer
to its root node. We'll use None
to represent the empty tree, just as we used None
for the empty linked list. In all leaf nodes, left and
right will be None.
Here is a small binary tree with just 3 nodes:
We can build this tree in Python as follows:
l = Node(7, None, None) r = Node(5, None, None) root = Node(4, l, r)
Here's a recursive function that counts the number of nodes in a binary tree:
def size(n):
if n == None: # empty tree
return 0
return 1 + size(n.left) + size(n.right)Let's try it on the tree we just built:
>>> size(root) 3
Recursion is a natural fit for trees, since the pattern of recursive calls in a function like this one can mirror the tree structure.
Here's a recursive function that computes the height of a binary tree:
def height(node):
if node == None:
return -1
return max(height(node.left), height(node.right)) + 1Let's try it:
>>> height(root) 1
The constant -1 in this function may seem surprising. Recall that the height of a binary tree is defined as the greatest depth of any node. If a binary tree has only a single node, then the depth of that node is 0, so the tree's height is 0. And so if a tree is empty, meaning that it has no nodes at all, then for consistency it must have a height of one less than that, or -1.
As another example, let's write a function that generates a complete binary tree of a given height h. Each node of the tree will contain a random value from 0 .. 99. Once more we will use recursion:
def complete(h):
if h == -1:
return None
left = complete(h - 1) # generate left subtree
right = complete(h - 1) # generate right subtree
return Node(randrange(100), left, right)Now we can easily generate a large tree:
>>> t = complete(10) >>> size(t) 2047 >>> height(t) 10
A binary search tree is a binary tree in which the values are ordered in a particular way that makes searching easy: for any node N with value v,
all values in N's left subtree are less than v
all values in N's right subtree are greater than v
For example, here is a binary search tree of integers:
We can use a binary search tree to store a set. To do this, we'll write a TreeSet class that holds the current root of a binary tree:
class TreeSet:
def __init__(self):
self.root = None
...It is not difficult to find whether a binary tree
contains a given value x. We begin
at the root. If the root's value is x, then we are done. Otherwise,
we compare x to the root's value v. If x < v, we move to the left
child; if x > v, we move to the right child. We proceed in this
way until we have found x or until we hit None,
in which case x is not in the tree.
Here's how we
can implement this in the TreeSet
class:
def contains(self, x):
n = self.root
while n != None:
if x == n.val:
return True
if x < n.val:
n = n.left
else:
n = n.right
return FalseInserting a value into a binary search tree is
also pretty straightforward. Beginning at the root, we look for an
insertion position, proceeding down the tree just as in the above
algorithm for contains. When we reach an empty left or
right child, we place the new node there. In the TreeSet
class:
# add a value, or do nothing if already present
def add(self, x):
n = Node(x, None, None) # new node to add
p = self.root
if p == None:
self.root = n
return
while True:
if x == p.val:
return # already present
elif x < p.val:
if p.left == None:
p.left = n
return
else:
p = p.left
else: # x > p.val
if p.right == None:
p.right = n
return
else:
p = p.rightDeleting a value from a binary search tree is a bit trickier. It's not hard to find the node to delete: we just walk down the tree, just like when searching or inserting. Once we've found the node N we want to delete, there are several cases.
If N is a leaf (it has no children), we can just remove it from the tree.
If N has only a single child, we replace N with its child. For example, we can delete node 15 in the binary tree above by replacing it with 18.
If N has two children, then we will replace
its value by the next largest value in the tree. To do this, we
start at N's right child and follow left child pointers for as long
as we can. This wil take us to the smallest node in N's right
subtree, which must be the next largest node in the tree after N.
Call this node M. We can easily remove M from the right subtree: M
has no left child, so we can remove it following either case (a) or
(b) above. Now we set N's value to the value that M had.
As
a concrete example, suppose that we want to delete the root node
(with value 10) in the tree above. This node has two children. We
start at its right child (20) and follow its left child pointer to
15. That’s as far as we can go in following left child pointers,
since 15 has no left child. So now we remove 15 (following case b
above), and then replace 10 with 15 at the root.
(Alternatively, we could delete node N by replacing its value with the next smallest value in the tree. That approach is symmetric and would work equally well. To find the next smallest value, we can start at N's left child and follow right child pointers for as long as we can.)
We won't give an implementation of this operation here, but writing this yourself is an excellent (and somewhat challenging) exercise.
It is not difficult to see that the add,
remove and contains operations described
above will all run in time O(h), where h is the height of a binary
search tree. What is their running time as a function of N, the
number of nodes in the tree?
First consider a complete binary search tree. As
we saw above, if a complete tree has N nodes then its height is
h = log2(N + 1) – 1 ≈ log2(N)
– 1 = O(log N). So add, remove, and
contains will all run in time O(log N).
Even if a tree is not complete, these operations will run in O(log N) time if the tree is not too tall given its number of nodes N, specfically if its height is O(log N). We call such a tree balanced.
Unfortunately not all binary trees are balanced. Suppose that we insert values into a binary search tree in ascending order:
t = TreeSet()
for i in range(1, 1000):
t.add(i)The tree will look like this:
This tree is completely unbalanced. It basically
looks like a linked list with an extra None
pointer at every node. add, remove and
contains will all run in O(N) on this tree.
How can we avoid an unbalanced tree such as this one? There are two possible ways. First, if we insert values into a binary search tree in a random order then that the tree will almost certainly be balanced. We will not prove this fact here (you might see a proof in the Algorithms and Data Structures class next semester).
Unfortunately it is not always practical to insert in a random order – for example, we may be reading a stream of values from a network and may need to insert each value as we receive it. So alternatively we can use a more advanced data structure known as a self-balancing binary tree, which automatically balances itself as values are inserted. Two examples of such structures are red‑black trees and AVL trees. We will not study these in this course, but you will see them in Algorithms and Data Structures next semester. For now, you should just be aware that they exist. In a self-balancing tree, the add(), remove() and contains() methods are all guaranteed to run in O(log N) time, for any possible sequence of tree operations.