Check Equivalence of Two Expression Trees in Python

Suppose, we are provided two expression trees. We have to write a program that checks the two expression trees and determines if the expression trees generate similar values. The two expression trees are provided to us in an in-order manner and we return a True value if they match, or else we return a False value.

So, if the input is like

then the output will be True.

The two expression trees evaluate to the same value.

To solve this, we will follow these steps:

• Define a function dfs() . This will take node, dic

  • if node is empty, then

    • return

  • if left of node and right of node is not empty, then

    • dic[value of node] := dic[value of node] + 1

  • dfs(left of node, dic)

  • dfs(right of node, dic)

• dic1 := a new map containing integer values

• dic2 := a new map containing integer values

• dfs(root1, dic1)

• dfs(root2, dic2)

• return True if dic1 is same as dic2.

Example

 Live Demo

import collections
class TreeNode:
   def __init__(self, val=0, left=None, right=None):
      self.val = val
      self.left = left
      self.right = right
def insert(temp,data):
   que = []
   que.append(temp)
   while (len(que)):
      temp = que[0]
      que.pop(0)
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data)
         else:
            temp.left = TreeNode(0)
         break
      else:
         que.append(temp.left)
      if (not temp.right):
         if data is not None:
            temp.right = TreeNode(data)
         else:
            temp.right = TreeNode(0)
         break
      else:
         que.append(temp.right)
def make_tree(elements):
   Tree = TreeNode(elements[0])
   for element in elements[1:]:
      insert(Tree, element)
   return Tree
def solve(root1, root2):
   dic1 = collections.defaultdict(int)
   dic2 = collections.defaultdict(int)
   def dfs(node, dic):
      if not node:
         return
      if not node.left and not node.right:
         dic[node.val] += 1
      dfs(node.left, dic)
      dfs(node.right, dic)
   dfs(root1, dic1)
   dfs(root2, dic2)
   return dic1 == dic2
root1 = make_tree([1, '+', 2, '*', 3, '+', 4 ])
root2 = make_tree([2, '+', 1, '*', 4, '+', 3])
print(solve(root1, root2))

Input

root1 = make_tree([1, '+', 2, '*', 3, '+', 4 ])
root2 = make_tree([2, '+', 1, '*', 4, '+', 3])

Output

True