5.1. Introduction to binary trees
(From Wikipedia, the free encyclopedia)
A binary search tree of size 9 and depth 3, with root 8 and leaves 1, 4, 7 and 13.
- Each node has a value.
- A total order is defined on these values.
- The left subtree of a node contains only values less than the node's value.
- The right subtree of a node contains only values greater than or equal to the node's value.
If a BST allows duplicate values, then it represents a multiset. This kind of tree uses non-strict inequalities. Everything in the left subtree of a node is strictly less than the value of the node, but everything in the right subtree is either greater than or equal to the value of the node.
If a BST doesn't allow duplicate values, then the tree represents a set with unique values, like the mathematical set. Trees without duplicate values use strict inequalities, meaning that the left subtree of a node only contains nodes with values that are less than the value of the node, and the right subtree only contains values that are greater.
The choice of storing equal values in the right subtree only is arbitrary; the left would work just as well. One can also permit non-strict equality in both sides. This allows a tree containing many duplicate values to be balanced better, but it makes searching more complex.
5.2. Operations
All operations on a binary tree make several calls to a
comparator, which is a
subroutine that computes the total order on any two values. In generic implementations of binary search trees, a program often provides a
callback to a comparator when it creates a tree, either explicitly or, in languages that support
type polymorphism, by having values be of a comparable type.
5.2.1. Searching
(From Wikipedia, the free encyclopedia)
Searching a binary tree for a specific value is a process that can be performed recursively because of the order in which values are stored. We begin by examining the root. If the value we are searching for equals the root, the value exists in the tree. If it is less than the root, then it must be in the left subtree, so we recursively search the left subtree in the same manner. Similarly, if it is greater than the root, then it must be in the right subtree, so we recursively search the right subtree. If we reach a leaf and have not found the value, then the item is not where it would be if it were present, so it does not lie in the tree at all. A comparison may be made with
binary search, which operates in nearly the same way but using random access on an array instead of following links.
def search_binary_tree(node, key):
if node is None:
return None # key not found
if key < node.key:
return search_binary_tree(node.left, key)
elif key > node.key:
return search_binary_tree(node.right, key)
else: # key is equal to node key
return node.value # found key
This operation requires
O(log n) time in the average case, but needs
O(n) time in the worst-case, when the unbalanced tree resembles a linked list.
5.2.2. Insertion
(From Wikipedia, the free encyclopedia)
Insertion begins as a search would begin; if the root is not equal to the value, we search the left or right subtrees as before. Eventually, we will reach an external node and add the value as its right or left child, depending on the node's value. In other words, we examine the root and recursively insert the new node to the left subtree if the new value is less than the root, or the right subtree if the new value is greater than or equal to the root.
Here's how a typical binary search tree insertion might be performed in
C++:
/* Inserts the node pointed to by "newNode" into the subtree rooted at "treeNode" */
void InsertNode(struct node *&treeNode, struct node *newNode)
{
if (treeNode == NULL)
treeNode = newNode;
else if (newNode->value < treeNode->value)
InsertNode(treeNode->left, newNode);
else
InsertNode(treeNode->right, newNode);
}
The above "destructive" procedural variant modifies the tree in place. It uses only constant space, but the previous version of the tree is lost. Alternatively, as in the following
Python example, we can reconstruct all ancestors of the inserted node; any reference to the original tree root remains valid, making the tree a
persistent data structure:
def binary_tree_insert(node, key, value):
if node is None:
return TreeNode(None, key, value, None)
if key == node.key:
return TreeNode(node.left, key, value, node.right)
if key < node.key:
return TreeNode(binary_tree_insert(node.left, key, value), node.key, node.value, node.right)
else:
return TreeNode(node.left, node.key, node.value, binary_tree_insert(node.right, key, value))
The part that is rebuilt uses Θ(log n) space in the average case and Ω(n) in the worst case (see
big-O notation).
In either version, this operation requires time proportional to the height of the tree in the worst case, which is
O(log n) time in the average case over all trees, but Ω(n2) time in the worst case.
Another way to explain insertion is that in order to insert a new node in the tree, its value is first compared with the value of the root. If its value is less than the root's, it is then compared with the value of the root's left child. If its value is greater, it is compared with the root's right child. This process continues, until the new node is compared with a leaf node, and then it is added as this node's right or left child, depending on its value.
There are other ways of inserting nodes into a binary tree, but this is the only way of inserting nodes at the leaves and at the same time preserving the BST structure.
5.2.3. Deletion
(From Wikipedia, the free encyclopedia)
There are several cases to be considered:
- Deleting a leaf: Deleting a node with no children is easy, as we can simply remove it from the tree.
- Deleting a node with one child: Delete it and replace it with its child.
- Deleting a node with two children: Suppose the node to be deleted is called N. We replace the value of N with either its in-order successor (the left-most child of the right subtree) or the in-order predecessor (the right-most child of the left subtree).
Once we find either the in-order successor or predecessor, swap it with N, and then delete it. Since both the successor and the predecessor must have fewer than two children, either one can be deleted using the previous two cases. In a good implementation, it is generally recommended to avoid consistently using one of these nodes, because this can
unbalance the tree.
Here is
C++ sample code for a destructive version of deletion. (We assume the node to be deleted has already been located using search.)
void DeleteNode(struct node * & node) {
if (node->left == NULL) {
struct node *temp = node;
node = node->right;
delete temp;
} else if (node->right == NULL) {
struct node *temp = node;
node = node->left;
delete temp;
} else {
// In-order predecessor (rightmost child of left subtree)
// Node has two children - get max of left subtree
struct node **temp = &node->left; // get left node of the original node
// find the rightmost child of the subtree of the left node
while ((*temp)->right != NULL) {
temp = &(*temp)->right;
}
// copy the value from the in-order predecessor to the original node
node->value = (*temp)->value;
// then delete the predecessor
DeleteNode(*temp);
}
}
Although this operation does not always traverse the tree down to a leaf, this is always a possibility; thus in the worst case it requires time proportional to the height of the tree. It does not require more even when the node has two children, since it still follows a single path and does not visit any node twice.
5.2.4. Traversal
(From Wikipedia, the free encyclopedia)
Compared to
linear data structures like
linked lists and one dimensional
arrays, which have only one logical means of traversal, tree structures can be traversed in many different ways. Starting at the root of a binary tree, there are three main steps that can be performed and the order in which they are performed define the traversal type. These steps are: Performing an action on the current node (referred to as "visiting" the node); or repeating the process with the subtrees rooted at our left and right children. Thus the process is most easily described through
recursion.
To traverse a non-empty binary tree in preorder, we perform the following three operations: 1. Visit the root. 2. Traverse the left subtree in preorder. 3. Traverse the right subtree in preorder.
To traverse a non-empty binary tree in inorder, perform the following operations: 1. Traverse the left subtree in inorder. 2. Visit the root. 3. Traverse the right subtree in inorder.
To traverse a non-empty binary tree in postorder, perform the following operations: 1. Traverse the left subtree in postorder. 2. Traverse the right subtree in postorder. 3. Visit the root. This is also called
Depth-first traversalFinally, trees can also be traversed in level-order, where we visit every node on a level before going to a lower level. This is also called
Breadth-first traversalOnce the binary search tree has been created, its elements can be retrieved
in order by recursively traversing the left subtree of the root node, accessing the node itself, then recursively traversing the right subtree of the node, continuing this pattern with each node in the tree as it's recursively accessed. The tree may also be traversed in
pre-order or
post-order traversals. The following is the implementation of these traversals:
preorder(node)
print node.value
if node.left ≠ null then preorder(node.left)
if node.right ≠ null then preorder(node.right)
inorder(node)
if node.left ≠ null then inorder(node.left)
print node.value
if node.right ≠ null then inorder(node.right)
postorder(node)
if node.left ≠ null then postorder(node.left)
if node.right ≠ null then postorder(node.right)
print node.value
All three sample implementations will require stack space proportional to the height of the tree. In a poorly balanced tree, this can be quite considerable.
5.2.5. Sort
(From Wikipedia, the free encyclopedia)
A binary search tree can be used to implement a simple but inefficient
sorting algorithm. Similarly to
heapsort, we insert all the values we wish to sort into a new ordered data structure — in this case a binary search tree — and then traverse it in order, building our result:
def build_binary_tree(values):
tree = None
for v in values:
tree = binary_tree_insert(tree, v)
return tree
def traverse_binary_tree(treenode):
if treenode is None: return []
else:
left, value, right = treenode
return (traverse_binary_tree(left) + [value] + traverse_binary_tree(right))
The worst-case time of build_binary_tree is Θ(n2) — if you feed it a sorted list of values, it chains them into a
linked list with no left subtrees. For example, build_binary_tree([1, 2, 3, 4, 5]) yields the tree (None, 1, (None, 2, (None, 3, (None, 4, (None, 5, None))))).
There are several schemes for overcoming this flaw with simple binary trees; the most common is the
self-balancing binary search tree. If this same procedure is done using such a tree, the overall worst-case time is
O(nlog n), which is
asymptotically optimal for a
comparison sort. In practice, the poor
cache performance and added overhead in time and space for a tree-based sort (particularly for node
allocation) make it inferior to other asymptotically optimal sorts such as
quicksort and
heapsort for static list sorting. On the other hand, it is one of the most efficient methods of incremental sorting, adding items to a list over time while keeping the list sorted at all times.