Alt hvad du behøver at vide om trædatastrukturer

Når du først lærer at kode, er det almindeligt at lære arrays som "hoveddatastruktur".

Til sidst lærer du hash tablesogså om. Hvis du forfølger en grad i datalogi, skal du tage en klasse om datastruktur. Du vil også lære om linked lists, queuesog stacks. Disse datastrukturer kaldes "lineære" datastrukturer, fordi de alle har en logisk start og en logisk slutning.

Når vi begynder at lære om, treesog graphsdet kan blive virkelig forvirrende. Vi gemmer ikke data på en lineær måde. Begge datastrukturer gemmer data på en bestemt måde.

Dette indlæg er til at hjælpe dig med at forstå trædatastrukturen bedre og afklare enhver forvirring, du måtte have om den.

I denne artikel lærer vi:

  • Hvad er et træ
  • Eksempler på træer
  • Dens terminologi, og hvordan det fungerer
  • Sådan implementeres træstrukturer i kode.

Lad os starte denne læringsrejse. :)

Definition

Når du starter programmeringen, er det almindeligt at forstå de lineære datastrukturer bedre end datastrukturer som træer og grafer.

Træer er velkendte som en ikke-lineær datastruktur. De lagrer ikke data på en lineær måde. De organiserer data hierarkisk.

Lad os dykke ned i virkelige eksempler!

Hvad mener jeg, når jeg siger en hierarkisk måde?

Forestil dig et stamtræ med forhold fra alle generationer: bedsteforældre, forældre, børn, søskende osv. Vi organiserer ofte stamtræer hierarkisk.

Ovenstående tegning er mit stamtræ. Tossico, Akikazu, Hitomi,og Takemier mine bedsteforældre.

Toshiakiog Juliana er mine forældre.

TK, Yuji, Bruno, og Kaioer mine forældres børn (mig og mine brødre).

En organisations struktur er et andet eksempel på et hierarki.

I HTML fungerer Document Object Model (DOM) som et træ.

Den HTMLtag indeholder andre tags. Vi har et headmærke og et bodymærke. Disse tags indeholder specifikke elementer. Den headtag har metaog titletags. Den bodytag har elementer, der viser i brugergrænsefladen, for eksempel h1, a, li, etc.

En teknisk definition

A treeer en samling enheder kaldet nodes. Noder er forbundet med edges. Hver nodeindeholder et valueeller data, og det kan eller ikke har et child node.

Den first nodeaf den treekaldes root. Hvis dette root nodeer forbundet af en anden node, rooter det så a parent nodeog det tilsluttede nodeer et child.

Alle Tree nodeser forbundet med kaldte links edges. Det er en vigtig del af trees, fordi det styrer forholdet mellem nodes.

Leaveser de sidste nodespå a tree.De er knuder uden børn. Ligesom virkelige træer, vi har den root, branchesog til sidst leaves.

Andre vigtige begreber at forstå er heightog depth.

Den heightaf a treeer længden af ​​den længste vej til en leaf.

Den depthaf a nodeer længden af ​​stien til dens root.

Terminologi resume

  • Root er den øverste nodeaftree
  • Edge er forbindelsen mellem tonodes
  • Barn er et, nodeder har enparent node
  • Forælder er en, nodeder har en edgetil enchild node
  • Blad er et node, der ikke har en child nodeitree
  • Højde er længden af ​​den længste vej til enleaf
  • Dybde er længden af ​​stien til densroot

Binære træer

Nu vil vi diskutere en bestemt type tree. Vi kalder det binary tree.

”I datalogi er et binært træ en treddatastruktur, hvor hver knudepunkt højst har to børn, der kaldes det venstre barn og det rigtige barn.” - Wikipedia

Så lad os se på et eksempel på en binary tree.

Lad os kode et binært træ

Den første ting, vi skal huske på, når vi implementerer en, binary treeer at det er en samling af nodes. Hver nodehar tre attributter: value, left_child, og right_child.

Hvordan implementerer vi en simpel, binary treeder initialiseres med disse tre egenskaber?

Lad os se.

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

Her er det. Vores binary treeklasse.

Når vi instantierer et objekt, videregiver vi value(nodens data) som en parameter. Se på left_childog right_child. Begge er indstillet til None.

Hvorfor?

For når vi skaber vores node, har den ingen børn. Vi har bare node data.

Lad os teste det:

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

Det er det.

Vi kan videregive string' a' som den valuetil vores Binary Tree node. Hvis vi udskriver value, left_childog right_childvi kan se værdierne.

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. :)

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:

To summarize the illustration of this tree:

  • anode will be the root of our binary Tree
  • aleft child is bnode
  • aright child is cnode
  • bright child is dnode (bnode doesn’t have a left child)
  • cleft child is enode
  • cright child is fnode
  • both e and fnodes 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.

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 rootnode 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

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

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.

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

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.

  1. First add the rootnode 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 rightchildren 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

“Et binært søgetræ kaldes undertiden ordnede eller sorterede binære træer, og det holder dets værdier i sorteret rækkefølge, så opslag og andre operationer kan bruge princippet om binær søgning” - Wikipedia

En vigtig egenskab ved a Binary Search Treeer, at værdien af ​​a Binary Search Treenodeer større end værdien af ​​dets afkom left child, men mindre end værdien af ​​dets afkom right child.”

Her er en oversigt over ovenstående illustration:

  • A er omvendt. De subtree7-5-8-6 behov for at være på den rigtige side, og de subtree2-1-3 behov for at være på den venstre.
  • B er den eneste korrekte mulighed. Det tilfredsstiller Binary Search Treeejendommen.
  • C har et problem: den nodemed værdien 4. Den skal være på venstre side af den, rootfordi den er mindre end 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.

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.

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 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 rightchildren, 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 rootnode. All the left subtreenodes will have smaller values than the rootnode. And all the right subtreenodes will have values greater than the rootnode.

Let’s take a look at an example.

Imagine that we have this tree.

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

Let’s break it down.

  1. We start with the rootnode 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.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 rootnode 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 (leafnode).
# |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 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

  • Introduction to Tree Data Structure by mycodeschool
  • Tree by Wikipedia
  • How To Not Be Stumped By Trees by the talented Vaidehi Joshi
  • Intro to Trees, Lecture by Professor Jonathan Cohen
  • Intro to Trees, Lecture by Professor David Schmidt
  • Intro to Trees, Lecture by Professor Victor Adamchik
  • Trees with Gayle Laakmann McDowell
  • Binary Tree Implementation and Tests by TK
  • Coursera Course: Data Structures by University of California, San Diego
  • Coursera Course: Data Structures and Performance by University of California, San Diego
  • Binary Search Tree concepts and Implementation by Paul Programming
  • Binær søgetræsimplementering og test af TK
  • Tree Traversal af Wikipedia
  • Binært søgetræ Fjern knudealgoritme af GeeksforGeeks
  • Binært søgetræ Fjern knudealgoritme af Algolist
  • At lære Python fra nul til helt