Binary search trees

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.

Here is the search algorithm in the Python programming language:

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.

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.

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).

Deletion

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.

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 traversal

Finally, 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 traversal

Once 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.

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.

Performance comparisons

D. A. Heger (2004) presented a performance comparison of binary search trees. Treap was found to have the best average performance, while red-black tree was found to have the smallest amount of performance fluctuations.

Optimal binary search trees

If we don't plan on modifying a search tree, and we know exactly how often each item will be accessed, we can construct an optimal binary search tree, which is a search tree where the average cost of looking up an item (the expected search cost) is minimized.

Assume that we know the elements and that for each element, we know the proportion of future lookups which will be looking for that element. We can then use a dynamic programming solution, detailed in section 15.5 of Introduction to Algorithms by Thomas H. Cormen Sec Edition, to construct the tree with the least possible expected search cost.

Even if we only have estimates of the search costs, such a system can considerably speed up lookups on average. For example, if you have a BST of English words used in a spell checker, you might balance the tree based on word frequency in text corpuses, placing words like "the" near the root and words like "agerasia" near the leaves. Such a tree might be compared with Huffman trees, which similarly seek to place frequently-used items near the root in order to produce a dense information encoding; however, Huffman trees only store data elements in leaves and these elements need not be ordered.

If we do not know the sequence in which the elements in the tree will be accessed in advance, we can use splay trees which are asymptotically as good as any static search tree we can construct for any particular sequence of lookup operations.

Alphabetic trees are Huffman trees with the additional constraint on order, or, equivalently, search trees with the modification that all elements are stored in the leaves. Faster algorithms exist for optimal alphabetic binary trees (OABTs).