Draw Binary Search Tree of a Word
Binary search trees explained
yourbasic.org
- Binary tree definitions
- Binary search tree
- Example
- Balanced trees with O(log n) time complexity
- Tree algorithms
- Inorder traversal
- Search
- Insert
Binary tree definitions
A binary tree is a data structure most easily described by recursion.
A binary tree
- is either empty,
- or consists of a node (also known as the root of the tree) and two subtrees, the left and right subtree, which are also binary trees.
A node with two empty subtrees is called aleaf.
If p
is a node and q
is the root of p
's subtree, we say that p
is the parent of q
and that q
is a child ofp
. Two nodes with the same parents are called siblings.
The set of ancestors if the noden
is defined recursively by these two rules:
-
n
is an ancestor ofn
. - If
q
is a child ofp
andq
is an ancestor ofn
, thenp
is an ancestor ofn
as well.
The set of ancestors to the noden
form a path fromn
to the root of the tree.
The depth of a node is the number of element on the path from the node to the root.
The depth of a tree is defined to be the largest depth of any node in the tree. The empty tree has depth 0.
Binary search tree
A binary search tree is a binary tree where each node contains a value from a well-ordered set.
For each node n
in a binary search tree the following invariants hold.
- Every node in the left subtree of
n
contains a value which is smaller than the value in the noden
. - Every node in the right subtree of
n
contains a value which is larger than the value in the noden
.
Example
This binary tree has 9 nodes and depth 4.
The root of the tree contains the value 8. The leaf values are 1, 4, 7 and 13.
- Each node in the left subtree has a value smaller than 8,
- while each node in the right subtree has a value larger than 8.
In fact, this is a binary search tree, since the corresponding invariant holds for each node in the tree.
Balanced trees with O(log n) time complexity
We say that a tree is well-balanced if each node in the tree has two subtrees with roughly the same number of nodes. It's possible to show that a well-balanced tree with n nodes has depth O(logn).
If we can manage to keep a binary search tree well-balanced, we get an ordered data structure with O(logn) worst-case time complexity for all basic operations: lookup, addition and removal.
There are several, more or less complicated, strategies to keep a binary search tree well-balanced.
- AVL trees came first,
- Red-black trees are used by Java's
TreeSet
, - Treaps, randomized binary search trees, are simple and elegant.
See the Treaps: randomized search trees article for a full description of treaps.
In this text we only present pseudocode for some basic operations on unbalanced binary search trees.
Warning: An unbalanced tree can be very inefficient. In the most extreme case, for example when all left subtrees are empty, the tree degrades into a singly linked list.
Tree algorithms
Inorder traversal
An inorder traversal of a binary search tree visits the nodes in sorted order.
// Visit all nodes of a binary search tree in sorted order. Algorithm inorder(root) if root is empty // do nothing else inorder(root.left) // do something with root inorder(root.right)
Search
It's pretty straightforward to implement the find operation in a binary search tree with iteration, but to keep things simple, here is a recursive version.
// Returns true if the value is found in the tree. Algorithm find(value, root) if root is empty return false if value = root.value return true if value < root.value return find(value, root.left) else return find(value, root.right)
Insert
To implement an algorithm that changes the structure of a tree, it's convenient to define a function that takes the root of the old tree as input, and returns the root of new updated tree.
// Adds a new node and returns the root of the updated tree. Algorithm insert(node, root) if root is empty return node if node.value = root.value // do nothing, element already in tree else if node.value < root.value root.left ← insert(node, root.left) else root.right ← insert(node, root.right) return root
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