Evaluate an expression binary tree - Daily Coding Challenge

Question

Here is my solution for the Daily Coding Challenge 50

Given an arithmetic expression in the form of a binary tree, write a function to evaluate it

Example

 *
 / \
 + +
 / \ / \
3 2 4 5
Return >> 45

Is this the most efficient way to do this?

"""Build an operation tree and calculate the result"""
class Tree:
 def __init__(self, value, left=None, right=None):
 """Initialise values"""
 self.value = value
 self.left = left
 self.right = right
 def determine_sum(self, total):
 if self.left == None and self.right == None:
 return int(self.value)
 if self.value == "*":
 total += self.left.determine_sum(total) * self.right.determine_sum(total)
 if self.value == "+":
 total += self.left.determine_sum(total) + self.right.determine_sum(total)
 if self.value == "-":
 total += self.left.determine_sum(total) - self.right.determine_sum(total)
 if self.value == "/":
 total += self.left.determine_sum(total) / self.right.determine_sum(total)
 return total
if __name__ == "__main__": #
 n = Tree("*")
 n.left = Tree("+")
 n.right = Tree("+")
 n.left.right = Tree("+")
 n.right.right = Tree("+")
 n.left.left = Tree("3")
 n.left.right.left = Tree("4")
 n.left.right.right = Tree("5")
 n.right.left = Tree("6")
 n.right.right.left = Tree("7")
 n.right.right.right = Tree("4")
 sum = n.determine_sum(0)
 print(sum)

• 200_success: In what sense is this a binary search tree?
• MrJoe: Sorry. Binary tree. My apologies

Answer 1

• dict> ifs
• operator contains all the functions you need.
• Taking total as an argument is unneeded.
• I would personally split the 'Tree' which is actually a Node into two types, operators and values. But that may go against the challenge.
• Use is to compare to None.

import operator
operators = {
 '*': operator.mul,
 '+': operator.add,
 '-': operator.sub,
 '/': operator.truediv,
}
class Tree:
 def __init__(self, value, left=None, right=None):
 """Initialise values"""
 self.value = value
 self.left = left
 self.right = right
 def determine(self):
 if self.left is None and self.right is None:
 return int(self.value)
 return operators[self.value](
 self.left.determine(),
 self.right.determine()
 )

• MrJoe: You never fail to amaze with with how succinctly you write code. Thanks
• Peilonrayz: @EML With time you'll be able to too :) You'll be able to see common pitfalls that you know how to fix without even thinking.