使用节点值而不是项号遍历非二叉树

Question

I want to create a non binary tree with 9 top level children, each of those children to have 8 children, each of those children to have 7 children, and so on..

I also want to be able to update data in these children by calling the appropriate node using it's value, a data item. Here is my code with comments:

import math

class NonBinTree:
    def __init__(self, value):
        self.value = value
        self.numTraverse = 0
        self.wins = 0
        self.UCB = 0
        self.nodes = []

    def add_node(self,value):
        self.nodes.append(NonBinTree(value))

    def read_numTraverse(self):
        return self.numTraverse

    def read_wins(self):
        return self.wins

    def read_value(self):
        return self.value

    def read_UCB(self):
        return self.UCB

    def update_number_times_traversed(self, data):
        self.numTraverse = self.numTraverse + data

    def update_wins(self, data):
        self.wins = self.wins + data

    def update_UCB(self):
        self.UCB = self.wins/self.numTraverse + (2*math.log(10,self.numTraverse)/self.numTraverse)

    def check_has_node(self, x):
        if self.value == x:
            return True
        else:
            y = False
            count = len(self.nodes)
            for i in range(count):
                if(self.nodes[i].read_value() == x):
                    y = True
                    break
            return y

    def __repr__(self):
        return f"NonBinTree({self.value}): {self.nodes}, {self.numTraverse}"

treeHead = NonBinTree(-1)

choices = [1,3,2]

for i in range(9):
    treeHead.add_node(i)

#the 0's here would be different numbers not all 0's
for i in range(9):
    for x in range(8):
        treeHead.nodes[i].add_node(0)

for i in range(9):
    for x in range(8):
        for y in range(7):
            treeHead.nodes[i].nodes[x].add_node(0)

# update tree after every choice
def update_tree(choices):
    count = len(choices)
    if(count==1):
        treeHead.nodes[choices[0]-1].update_number_times_traversed(1)
    elif(count==2):
        if(choices[1]>choices[0]):
            treeHead.nodes[choices[0] - 1].nodes[choices[1] - 2].update_number_times_traversed(1)
        else:
            treeHead.nodes[choices[0] - 1].nodes[choices[1] - 1].update_number_times_traversed(1)
    elif(count==3):
        if(choices[2]>choices[1]):
            if (choices[1] > choices[0]):
                treeHead.nodes[choices[0] - 1].nodes[choices[1] - 2].nodes[choices[2] - 2].update_number_times_traversed(1)
            else:
                if (choices[2] > choices[0]):
                    treeHead.nodes[choices[0] - 1].nodes[choices[1] - 1].nodes[choices[2] - 2].update_number_times_traversed(1)
                else:
                    treeHead.nodes[choices[0] - 1].nodes[choices[1] - 1].nodes[choices[2] - 1].update_number_times_traversed(1)
        else:
            if (choices[0] > choices[1]):
                treeHead.nodes[choices[0] - 1].nodes[choices[1] - 1].nodes[choices[2] - 1].update_number_times_traversed(1)
            else:
                if (choices[2] > choices[0]):
                    treeHead.nodes[choices[0] - 1].nodes[choices[1] - 2].nodes[choices[2] - 1].update_number_times_traversed(1)
                else:
                    treeHead.nodes[choices[0] - 1].nodes[choices[1] - 2].nodes[choices[2] - 2].update_number_times_traversed(1)

Instead of an update like treeHead.nodes[choices[0]-1].update_number_times_traversed(1) I would like to update using the value of a node, for example in pseudo code: treeHead.nodes[value:1].nodes[value:19].update_number_times_traversed(1) Is this possible?

Answer 1

for this self.nodes = [] cannot be a list, but instead a dict:

{
  <value_1>: <node1>,
  <value_2>: <node2>,
  <value_3>: <node3>,
}

You can define it like:

self.nodes: Dict[int, NonBinTree] = {}

or

self.nodes = defaultdict(NonBinTree)

to add a node you just do:

   def add_node(self, value):
        self.nodes[value] = NonBinTree(value)

WARNING : value will have to be unique, otherwise you'd overwrite another node

• treeHead.nodes[value:1] syntax won't work though.
• treeHead.nodes[value].nodes will still work

If you want to do more complex things, self.nodes would need to be an instance of a data structure class you define, allowing slicing, index based access, etc etc