a simple implementation of a Binary Search Tree in Python

binarySearchTree.py

class Node:
    def __init__(self, val):
        self.val = val
        self.leftChild = None
        self.rightChild = None
    
    def get(self):
        return self.val
    
    def set(self, val):
        self.val = val
        
    def getChildren(self):
        children = []
        if(self.leftChild != None):
            children.append(self.leftChild)
        if(self.rightChild != None):
            children.append(self.rightChild)
        return children
        
class BST:
    def __init__(self):
        self.root = None

    def setRoot(self, val):
        self.root = Node(val)

    def insert(self, val):
        if(self.root is None):
            self.setRoot(val)
        else:
            self.insertNode(self.root, val)

    def insertNode(self, currentNode, val):
        if(val <= currentNode.val):
            if(currentNode.leftChild):
                self.insertNode(currentNode.leftChild, val)
            else:
                currentNode.leftChild = Node(val)
        elif(val > currentNode.val):
            if(currentNode.rightChild):
                self.insertNode(currentNode.rightChild, val)
            else:
                currentNode.rightChild = Node(val)

    def find(self, val):
        return self.findNode(self.root, val)

    def findNode(self, currentNode, val):
        if(currentNode is None):
            return False
        elif(val == currentNode.val):
            return True
        elif(val < currentNode.val):
            return self.findNode(currentNode.leftChild, val)
        else:
            return self.findNode(currentNode.rightChild, val)

Thanks for this implementation! :)

Just two comments:

1. Python uses underscore as method and variable naming: find_node, left_children, ...

2. Maybe the setRoot needs to be a private method. If we insert 5 new nodes, and then call setRoot method, it will clear all nodes we inserted. It is not a problem we want to handle.

Great work! 👍

Straight forward implementation. However, the insert function shall be updated to ignore adding redundant values.

can you make delete function
part for this

Thanks!

Thanks, it helped me.

Thank you very much

Could you please show how deletion is done?

For deletion, the function in Node class would be

def delete(self):
        # cases where the node is a leaf, just remove it
        if self.leftNode == None and self.rightNode == None:
            if self.parent.value > self.value:
                self.parent.rightNode = None
            else:
                self.parent.leftNode = None
            # remove parent's Link
            self.parent = None
            return

        # cases where children are only in one direction, just change the pointers
        # where children only in left direction
        if self.rightNode == None:
            self.parent.leftNode = self.leftNode
            self.leftNode.parent = self.parent

            self.parent = None
            self.leftNode = None
            return
        # where children only in rightdirection
        elif self.leftNode == None:
            self.parent.rightNode = self.rightNode
            self.rightNode.parent = self.parent

            self.parent = None
            self.leftNode = None
            return

        
        # cases where there are children in both the directions
        # swap values between currentNode and its succesorNode
        # delete the successor_node(which contains the value to be removed from tree), note that the successor_node is a leaf
        successor_node = self.rightNode.find_min_in_sub_tree()
        successor_node.value, self.value = self.value, successor_node.value
        successor_node.delete()

not easy part: why self.rightNode.find_min_in_sub_tree() was done. The reason behind doing this is to find its next biggest value(its successor) in the tree

the implementaion of find_min_in_sub_tree. This function finds the minimum element in the tree below it, by continuously looking at the left

def find_min_in_sub_tree(self):
        if self.leftNode == None:
            return self
        
        if self.leftNode != None:
            return self.leftNode.find_min_in_sub_tree()

Reference

@shree675, @manjitha27
Do let me know if something is wrong

EDIT- Thanks to @Aniket2ten96 for pointing out the mistake

@rakaar I think the code to find the successor_node might not yield the correct result. Suppose a bst rooted at 69, and its right side nodes are 70,82,84. Now I want to find 69's successor, according to your code i will go to right and i will find the node with maximum key and that will be the successor of 69. In the above example the maximum key is 84, which is not the successor of 69(successor of 69 is 70 in the above example) . I think to find the successor, you have to go right and find the node with minimum key value, which is in the above example 70. Correct me if I am wrong

Thanks a lot @Aniket2ten96 for pointing it out
I have made the changes
You are right, it should be finding out the minimum in the subtree where self.rightNode acts as the root. Hence replaced the wrong function with find_min_in_sub_tree along with its implementation

Thank you for the clean example!