When you first learn to code, it’s common to learn arrays as the “main data structure.”

Eventually, you will learn about hash tables too. If you are pursuing a Computer Science degree, you have to take a class on data structure. You will also learn about linked lists, queues, and stacks. Those data structures are called “linear” data structures because they all have a logical start and a logical end.

When we start learning about trees and graphs, it can get really confusing. We don’t store data in a linear way. Both data structures store data in a specific way.

This post is to help you better understand the Tree Data Structure and to clarify any confusion you may have about it.

In this article, we will learn:

  • What is a tree
  • Examples of trees
  • Its terminology and how it works
  • How to implement tree structures in code.

Let’s start this learning journey. :)

Definition

When starting out programming, it is common to understand better the linear data structures than data structures like trees and graphs.

Trees are well-known as a non-linear data structure. They don’t store data in a linear way. They organize data hierarchically.

Let’s dive into real life examples!

What do I mean when I say in a hierarchical way?

Imagine a family tree with relationships from all generation: grandparents, parents, children, siblings, etc. We commonly organize family trees hierarchically.

1*MasdC5DmucEU2abIXQe45Q
My family tree

The above drawing is is my family tree. Tossico, Akikazu, Hitomi, and Takemi are my grandparents.

Toshiaki and Juliana are my parents.

TK, Yuji, Bruno, and Kaio are the children of my parents (me and my brothers).

An organization’s structure is another example of a hierarchy.

1*GsBCmW5E1GuJ3MpH3Zz0Ew
A company’s structure is an example of a hierarchy

In HTML, the Document Object Model (DOM) works as a tree.

1*dLXUdR4NuIZG8GJdu_Cinw
Document Object Model (DOM)

The HTML tag contains other tags. We have a head tag and a body tag. Those tags contains specific elements. The head tag has meta and title tags. The body tag has elements that show in the user interface, for example, h1, a, li, etc.

A technical definition

A tree is a collection of entities called nodes. Nodes are connected by edges. Each node contains a value or data, and it may or may not have a child node .

1*3WN7tIQ-kNBQmY9MgvTuOA

The first node of the tree is called the root. If this root node is connected by another node, the root is then a parent node and the connected node is a child.

1*9AtR3bhhlMJxQlaUVEQgrw

All Tree nodes are connected by links called edges. It’s an important part of trees, because it’s manages the relationship between nodes.

1*j5qKwIxKcEjoxy88EOc1Rg

Leaves are the last nodes on a tree. They are nodes without children. Like real trees, we have the root, branches, and finally the leaves.

1*c9_5uMUsIy4Q3OA7Q8bJiw

Other important concepts to understand are height and depth.

The height of a tree is the length of the longest path to a leaf.

The depth of a node is the length of the path to its root.

Terminology summary

  • Root is the topmost node of the tree
  • Edge is the link between two nodes
  • Child is a node that has a parent node
  • Parent is a node that has an edge to a child node
  • Leaf is a node that does not have a child node in the tree
  • Height is the length of the longest path to a leaf
  • Depth is the length of the path to its root

Binary trees

Now we will discuss a specific type of tree. We call it thebinary tree.

“In computer science, a binary tree is a tree data structure in which each node has at the most two children, which are referred to as the left child and the right child.” — Wikipedia

So let’s look at an example of a binary tree.

1*ofbwuz4inpf2OlB-l9gtHw

Let’s code a binary tree

The first thing we need to keep in mind when we implement a binary tree is that it is a collection of nodes. Each node has three attributes: value, left_child, and right_child.

How do we implement a simple binary tree that initializes with these three properties?

Let’s take a look.

class BinaryTree:
    def __init__(self, value):
        self.value = value
        self.left_child = None
        self.right_child = None

Here it is. Our binary tree class.

When we instantiate an object, we pass the value (the data of the node) as a parameter. Look at the left_child and the right_child. Both are set to None.

Why?

Because when we create our node, it doesn’t have any children. We just have the node data.

Let’s test it:

tree = BinaryTree('a')
print(tree.value) # a
print(tree.left_child) # None
print(tree.right_child) # None

That’s it.

We can pass the stringa’ as the value to our Binary Tree node. If we print the value, left_child, and right_child, we can see the values.

Let’s go to the insertion part. What do we need to do here?

We will implement a method to insert a new node to the right and to the left.

Here are the rules:

  • If the current node doesn’t have a left child, we just create a new nodeand set it to the current node’s left_child.
  • If it does have the left child, we create a new node and put it in the current left child’s place. Allocate this left child node to the new node’s left child.

Let’s draw it out. :)

1*ofbwuz4inpf2OlB-l9gtHw

Here’s the code:

def insert_left(self, value):
    if self.left_child == None:
        self.left_child = BinaryTree(value)
    else:
        new_node = BinaryTree(value)
        new_node.left_child = self.left_child
        self.left_child = new_node

Again, if the current node doesn’t have a left child, we just create a new node and set it to the current node’s left_child. Or else we create a new node and put it in the current left child’s place. Allocate this left child node to the new node’s left child.

And we do the same thing to insert a right child node.

def insert_right(self, value):
    if self.right_child == None:
        self.right_child = BinaryTree(value)
    else:
        new_node = BinaryTree(value)
        new_node.right_child = self.right_child
        self.right_child = new_node

Done. :)

But not entirely. We still need to test it.

Let’s build the followingtree:

1*V_EUgNXVc8Wy9H1-JoqT3g

To summarize the illustration of this tree:

  • a node will be the root of our binary Tree
  • a left child is b node
  • a right child is c node
  • b right child is d node (b node doesn’t have a left child)
  • c left child is e node
  • c right child is f node
  • both e and f nodes do not have children

So here is the code for the tree:

a_node = BinaryTree('a')
a_node.insert_left('b')
a_node.insert_right('c')

b_node = a_node.left_child
b_node.insert_right('d')

c_node = a_node.right_child
c_node.insert_left('e')
c_node.insert_right('f')

d_node = b_node.right_child
e_node = c_node.left_child
f_node = c_node.right_child

print(a_node.value) # a
print(b_node.value) # b
print(c_node.value) # c
print(d_node.value) # d
print(e_node.value) # e
print(f_node.value) # f

Insertion is done.

Now we have to think about tree traversal.

We have two options here: Depth-First Search (DFS) and Breadth-First Search (BFS).

  • DFS “is an algorithm for traversing or searching tree data structure. One starts at the root and explores as far as possible along each branch before backtracking.”Wikipedia
  • BFS “is an algorithm for traversing or searching tree data structure. It starts at the tree root and explores the neighbor nodes first, before moving to the next level neighbors.”Wikipedia

So let’s dive into each tree traversal type.

Depth-First Search (DFS)

DFS explores a path all the way to a leaf before backtracking and exploring another path. Let’s take a look at an example with this type of traversal.

1*-sCuUx3R9e1ougu2pGdThg

The result for this algorithm will be 1–2–3–4–5–6–7.

Why?

Let’s break it down.

  1. Start at the root (1). Print it.

2. Go to the left child (2). Print it.

3. Then go to the left child (3). Print it. (This node doesn’t have any children)

4. Backtrack and go the right child (4). Print it. (This node doesn’t have any children)

5. Backtrack to the root node and go to the right child (5). Print it.

6. Go to the left child (6). Print it. (This node doesn’t have any children)

7. Backtrack and go to the right child (7). Print it. (This node doesn’t have any children)

8. Done.

When we go deep to the leaf and backtrack, this is called DFS algorithm.

Now that we are familiar with this traversal algorithm, we will discuss types of DFS: pre-order, in-order, and post-order.

Pre-order

This is exactly what we did in the above example.

  1. Print the value of the node.
  2. Go to the left child and print it. This is if, and only if, it has a left child.
  3. Go to the right child and print it. This is if, and only if, it has a right child.
def pre_order(self):
    print(self.value)

    if self.left_child:
        self.left_child.pre_order()

    if self.right_child:
        self.right_child.pre_order()

In-order

1*-sCuUx3R9e1ougu2pGdThg

The result of the in-order algorithm for this tree example is 3–2–4–1–6–5–7.

The left first, the middle second, and the right last.

Now let’s code it.

def in_order(self):
    if self.left_child:
        self.left_child.in_order()

    print(self.value)

    if self.right_child:
        self.right_child.in_order()
  1. Go to the left child and print it. This is if, and only if, it has a left child.
  2. Print the node’s value
  3. Go to the right child and print it. This is if, and only if, it has a right child.

Post-order

1*-sCuUx3R9e1ougu2pGdThg

The result of the post order algorithm for this tree example is 3–4–2–6–7–5–1.

The left first, the right second, and the middle last.

Let’s code this.

def post_order(self):
    if self.left_child:
        self.left_child.post_order()

    if self.right_child:
        self.right_child.post_order()

    print(self.value)
  1. Go to the left child and print it. This is if, and only if, it has a left child.
  2. Go to the right child and print it. This is if, and only if, it has a right child.
  3. Print the node’s value

Breadth-First Search (BFS)

BFS algorithm traverses the tree level by level and depth by depth.

1*ZNxp_NkRZLCeak85rreebA

Here is an example that helps to better explain this algorithm:

1*-sCuUx3R9e1ougu2pGdThg

So we traverse level by level. In this example, the result is 1–2–5–3–4–6–7.

  • Level/Depth 0: only node with value 1
  • Level/Depth 1: nodes with values 2 and 5
  • Level/Depth 2: nodes with values 3, 4, 6, and 7

Now let’s code it.

def bfs(self):
    queue = Queue()
    queue.put(self)

    while not queue.empty():
        current_node = queue.get()
        print(current_node.value)

        if current_node.left_child:
            queue.put(current_node.left_child)

        if current_node.right_child:
            queue.put(current_node.right_child)

To implement a BFS algorithm, we use the queue data structure to help.

How does it work?

Here’s the explanation.

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  1. First add the root node into the queue with the put method.
  2. Iterate while the queue is not empty.
  3. Get the first node in the queue, and then print its value.
  4. Add both left and right children into the queue (if the current nodehas children).
  5. Done. We will print the value of each node, level by level, with our queuehelper.

Binary Search tree

“A Binary Search Tree is sometimes called ordered or sorted binary trees, and it keeps its values in sorted order, so that lookup and other operations can use the principle of binary search” — Wikipedia

An important property of a Binary Search Tree is that the value of a Binary Search Tree nodeis larger than the value of the offspring of its left child, but smaller than the value of the offspring of its right child.

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Here is a breakdown of the above illustration:

  • A is inverted. The subtree 7–5–8–6 needs to be on the right side, and the subtree 2–1–3 needs to be on the left.
  • B is the only correct option. It satisfies the Binary Search Tree property.
  • C has one problem: the node with the value 4. It needs to be on the left side of the root because it is smaller than 5.

Let’s code a Binary Search Tree!

Now it’s time to code!

What will we see here? We will insert new nodes, search for a value, delete nodes, and the balance of the tree.

Let’s start.

Insertion: adding new nodes to our tree

Imagine that we have an empty tree and we want to add new nodes with the following values in this order: 50, 76, 21, 4, 32, 100, 64, 52.

The first thing we need to know is if 50 is the root of our tree.

1*fxSlTwgQSN_DlzfEmcxqQg

We can now start inserting node by node.

  • 76 is greater than 50, so insert 76 on the right side.
  • 21 is smaller than 50, so insert 21 on the left side.
  • 4 is smaller than 50. Node with value 50 has a left child 21. Since 4 is smaller than 21, insert it on the left side of this node.
  • 32 is smaller than 50. Node with value 50 has a left child 21. Since 32 is greater than 21, insert 32 on the right side of this node.
  • 100 is greater than 50. Node with value 50 has a right child 76. Since 100 is greater than 76, insert 100 on the right side of this node.
  • 64 is greater than 50. Node with value 50 has a right child 76. Since 64 is smaller than 76, insert 64 on the left side of this node.
  • 52 is greater than 50. Node with value 50 has a right child 76. Since 52 is smaller than 76, node with value 76 has a left child 64. 52 is smaller than 64, so insert 54 on the left side of this node.
1*LlLDNx7wgJfH6VAGnyAbIQ

Do you notice a pattern here?

Let’s break it down.

  1. Is the new node value greater or smaller than the current node?
  2. If the value of the new node is greater than the current node, go to the right subtree. If the current node doesn’t have a right child, insert it there, or else backtrack to step #1.
  3. If the value of the new node is smaller than the current node, go to the left subtree. If the current node doesn’t have a left child, insert it there, or else backtrack to step #1.
  4. We did not handle special cases here. When the value of a new node is equal to the current value of the node, use rule number 3. Consider inserting equal values to the left side of the subtree.

Now let’s code it.

class BinarySearchTree:
    def __init__(self, value):
        self.value = value
        self.left_child = None
        self.right_child = None

    def insert_node(self, value):
        if value <= self.value and self.left_child:
            self.left_child.insert_node(value)
        elif value <= self.value:
            self.left_child = BinarySearchTree(value)
        elif value > self.value and self.right_child:
            self.right_child.insert_node(value)
        else:
            self.right_child = BinarySearchTree(value)

It seems very simple.

The powerful part of this algorithm is the recursion part, which is on line 9 and line 13. Both lines of code call the insert_node method, and use it for its left and right children, respectively. Lines 11 and 15 are the ones that do the insertion for each child.

Let’s search for the node value… Or not…

The algorithm that we will build now is about doing searches. For a given value (integer number), we will say if our Binary Search Tree does or does not have that value.

An important item to note is how we defined the tree insertion algorithm. First we have our root node. All the left subtree nodes will have smaller values than the root node. And all the right subtree nodes will have values greater than the root node.

Let’s take a look at an example.

Imagine that we have this tree.

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Now we want to know if we have a node based on value 52.

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Let’s break it down.

  1. We start with the root node as our current node. Is the given value smaller than the current node value? If yes, then we will search for it on the left subtree.
  2. Is the given value greater than the current node value? If yes, then we will search for it on the right subtree.
  3. If rules #1 and #2 are both false, we can compare the current node value and the given value if they are equal. If the comparison returns true, then we can say, “Yeah! Our tree has the given value,” otherwise, we say, “Nooo, it hasn’t.”

Now let’s code it.

class BinarySearchTree:
    def __init__(self, value):
        self.value = value
        self.left_child = None
        self.right_child = None

    def find_node(self, value):
        if value < self.value and self.left_child:
            return self.left_child.find_node(value)
        if value > self.value and self.right_child:
            return self.right_child.find_node(value)

        return value == self.value

Let’s beak down the code:

  • Lines 8 and 9 fall under rule #1.
  • Lines 10 and 11 fall under rule #2.
  • Line 13 falls under rule #3.

How do we test it?

Let’s create our Binary Search Tree by initializing the root node with the value 15.

bst = BinarySearchTree(15)

And now we will insert many new nodes.

bst.insert_node(10)
bst.insert_node(8)
bst.insert_node(12)
bst.insert_node(20)
bst.insert_node(17)
bst.insert_node(25)
bst.insert_node(19)

For each inserted node , we will test if our find_node method really works.

print(bst.find_node(15)) # True
print(bst.find_node(10)) # True
print(bst.find_node(8)) # True
print(bst.find_node(12)) # True
print(bst.find_node(20)) # True
print(bst.find_node(17)) # True
print(bst.find_node(25)) # True
print(bst.find_node(19)) # True

Yeah, it works for these given values! Let’s test for a value that doesn’t exist in our Binary Search Tree.

print(bst.find_node(0)) # False

Oh yeah.

Our search is done.

Deletion: removing and organizing

Deletion is a more complex algorithm because we need to handle different cases. For a given value, we need to remove the node with this value. Imagine the following scenarios for this node : it has no children, has a single child, or has two children.

  • Scenario #1: A node with no children (leaf node).
#        |50|                              |50|
#      /      \                           /    \
#    |30|     |70|   (DELETE 20) --->   |30|   |70|
#   /    \                                \
# |20|   |40|                             |40|

If the node we want to delete has no children, we simply delete it. The algorithm doesn’t need to reorganize the tree.

  • Scenario #2: A node with just one child (left or right child).
#        |50|                              |50|
#      /      \                           /    \
#    |30|     |70|   (DELETE 30) --->   |20|   |70|
#   /            
# |20|

In this case, our algorithm needs to make the parent of the node point to the child node. If the node is the left child, we make the parent of the left child point to the child. If the node is the right child of its parent, we make the parent of the right child point to the child.

  • Scenario #3: A node with two children.
#        |50|                              |50|
#      /      \                           /    \
#    |30|     |70|   (DELETE 30) --->   |40|   |70|
#   /    \                             /
# |20|   |40|                        |20|

When the node has 2 children, we need to find the node with the minimum value, starting from the node’sright child. We will put this node with minimum value in the place of the node we want to remove.

It’s time to code.

def remove_node(self, value, parent):
    if value < self.value and self.left_child:
        return self.left_child.remove_node(value, self)
    elif value < self.value:
        return False
    elif value > self.value and self.right_child:
        return self.right_child.remove_node(value, self)
    elif value > self.value:
        return False
    else:
        if self.left_child is None and self.right_child is None and self == parent.left_child:
            parent.left_child = None
            self.clear_node()
        elif self.left_child is None and self.right_child is None and self == parent.right_child:
            parent.right_child = None
            self.clear_node()
        elif self.left_child and self.right_child is None and self == parent.left_child:
            parent.left_child = self.left_child
            self.clear_node()
        elif self.left_child and self.right_child is None and self == parent.right_child:
            parent.right_child = self.left_child
            self.clear_node()
        elif self.right_child and self.left_child is None and self == parent.left_child:
            parent.left_child = self.right_child
            self.clear_node()
        elif self.right_child and self.left_child is None and self == parent.right_child:
            parent.right_child = self.right_child
            self.clear_node()
        else:
            self.value = self.right_child.find_minimum_value()
            self.right_child.remove_node(self.value, self)

        return True
  1. First: Note the parameters value and parent. We want to find the nodethat has this value , and the node’s parent is important to the removal of the node.
  2. Second: Note the returning value. Our algorithm will return a boolean value. It returns True if it finds the node and removes it. Otherwise it will return False.
  3. From line 2 to line 9: We start searching for the node that has the valuethat we are looking for. If the value is smaller than the current nodevalue , we go to the left subtree, recursively (if, and only if, the current node has a left child). If the value is greater, go to the right subtree, recursively.
  4. Line 10: We start to think about the remove algorithm.
  5. From line 11 to line 13: We cover the node with no children , and it is the left child from its parent. We remove the node by setting the parent’s left child to None.
  6. Lines 14 and 15: We cover the node with no children , and it is the right child from it’s parent. We remove the node by setting the parent’s right child to None.
  7. Clear node method: I will show the clear_node code below. It sets the nodes left child , right child, and its value to None.
  8. From line 16 to line 18: We cover the node with just one child (left child), and it is the left child from it’s parent. We set the parent's left child to the node’s left child (the only child it has).
  9. From line 19 to line 21: We cover the node with just one child (left child), and it is the right child from its parent. We set the parent's right child to the node’s left child (the only child it has).
  10. From line 22 to line 24: We cover the node with just one child (right child), and it is the left child from its parent. We set the parent's left child to the node’s right child (the only child it has).
  11. From line 25 to line 27: We cover the node with just one child (right child) , and it is the right child from its parent. We set the parent's right child to the node’s right child (the only child it has).
  12. From line 28 to line 30: We cover the node with both left and rightchildren. We get the node with the smallest value (the code is shown below) and set it to the value of the current node . Finish it by removing the smallest node.
  13. Line 32: If we find the node we are looking for, it needs to return True. From line 11 to line 31, we handle this case. So just return True and that’s it.
  • To use the clear_node method: set the None value to all three attributes — (value, left_child, and right_child)
def clear_node(self):
    self.value = None
    self.left_child = None
    self.right_child = None
  • To use the find_minimum_value method: go way down to the left. If we can’t find anymore nodes, we found the smallest one.
def find_minimum_value(self):
    if self.left_child:
        return self.left_child.find_minimum_value()
    else:
        return self.value

Now let’s test it.

We will use this tree to test our remove_node algorithm.

#        |15|
#      /      \
#    |10|     |20|
#   /    \    /    \
# |8|   |12| |17| |25|
#              \
#              |19|

Let’s remove the node with the value 8. It’s a node with no child.

print(bst.remove_node(8, None)) # True
bst.pre_order_traversal()

#     |15|
#   /      \
# |10|     |20|
#    \    /    \
#   |12| |17| |25|
#          \
#          |19|

Now let’s remove the node with the value 17. It’s a node with just one child.

print(bst.remove_node(17, None)) # True
bst.pre_order_traversal()

#        |15|
#      /      \
#    |10|     |20|
#       \    /    \
#      |12| |19| |25|

Finally, we will remove a node with two children. This is the root of our tree.

print(bst.remove_node(15, None)) # True
bst.pre_order_traversal()

#        |19|
#      /      \
#    |10|     |20|
#        \        \
#        |12|     |25|

The tests are now done. :)

That’s all for now!

We learned a lot here.

Congrats on finishing this dense content. It’s really tough to understand a concept that we do not know. But you did it. :)

This is one more step forward in my journey to learning and mastering algorithms and data structures. You can see the documentation of my complete journey here on my Renaissance Developer publication.

Have fun, keep learning and coding.

My Twitter & Github. ☺

Additional resources