在 Python 中迭代删除二叉搜索树中的值

Question

I'm currently taking a course where I'm learning data structures and algorithms, and I'm learning about BST's. I already got the code to work and understand most of it, but I got a question for the deletion function. This is what my code looks like:

class BST:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

    def insert(self, value):
        currentNode = self

        while True:
            if value < currentNode.value:
                if currentNode.left is None:
                    currentNode.left = BST(value)
                    break
                else:
                    currentNode = currentNode.left
            else:
                if currentNode.right is None:
                    currentNode.right = BST(value)
                    break
                else:
                    currentNode = currentNode.right
        return self

    def contains(self, value):
        currentNode = self

        while currentNode is not None:
            if value < currentNode.value:
                currentNode = currentNode.left
            elif value > currentNode.value:
                currentNode = currentNode.right
            else:
                return True
        return False

    def remove(self, value, parentNode = None):
        currentNode = self

        while currentNode is not None:
            if value < currentNode.value:
                parentNode = currentNode
                currentNode = currentNode.left  
            elif value > currentNode.value:
                parentNode = currentNode
                currentNode = currentNode.right
            #Found the node
            else:   
            #two child ondes
                if currentNode.left is not None and currentNode.right is not None:
                    currentNode.value = currentNode.right.getMinValue()      #get the left number from the right subtree
                    currentNode.right.remove(currentNode.value, currentNode) #remove that most left number by using remove() 
                                                                             #on the right currentNode
                #root node
                elif parentNode is None:
                    if currentNode.left is not None:
                        currentNode.value = currentNode.left.value
                        currentNode.right = currentNode.left.right
                        currentNode.left = currentNode.left.left
                    elif currentNode.right is not None:
                        currentNode.value = currentNode.right.value
                        currentNode.left = currentNode.right.left
                        currentNode.right = currentNode.right.right
                    #only 1 item left in BST
                    else:
                        pass
                #one child node
                elif parentNode.left == currentNode:
                    parentNode.left = currentNode.left if currentNode.left is not None else currentNode.right
                elif parentNode.right == currentNode:
                    parentNode.right = currentNode.left if currentNode.left is not None else currentNode.right
                break
        return self

    def getMinValue(self):
        currentNode = self

        while currentNode.left is not None:
            currentNode = currentNode.left
        return currentNode.value

I understand that, for the delete function:

1. The while loop will iterate over each node and run until there are no more nodes
2. The first if and elif are used to find the node you are trying to remove.
3. The else works for the actual deletion, which has 3 differnt options: Either currentNode has two childs, and you just replace it by the leftmost value of the right node and delete this leftmost value from the right noe. The other case is that parentNode has no parent, which would be the root Node case. And the last case is when you have only one child node all you have to do is change the value of currentNode either to its left node or right node (depending which one it has).

What I don't clearly understand is the logic behind the conditions, and how it works when we want to delete a root node. Isn't the code supposed to also run the first condition, which is the one for two child nodes? I'm almost certain that this isn't supposed to happen, and the condition should only run for its special case. I have seen again and again the video explanation, but I just can't get a hang of it.

Answer 1

when we want to delete a root node. Isn't the code supposed to also run the first condition, which is the one for two child nodes?

Even in case the root node must be removed, it actually does evaluate the first condition. If the root node has both a left and right child, then "option 1" applies to it: that first option deals fine with any node that has two children, whether it be a root node or not. No distinction needs to be made between a root or non-root node in this option.

The other two options are only for nodes that do not have two children. You seem to suggest (also in the code comments) that only option 3 deals with that case, but also option 2 does. Option 2 is for when the node node does not have two children and it is the root. If the root would have had 2 children, it would be treated as option 1.