Medium
Check If Two Expression Trees are Equivalent — Python
Full explanation · Time O(n) · Space O(1)
# Time: O(n)
# Space: O(1)
import collections
import functools
# Definition for a binary tree node.
class Node(object):
def __init__(self, val=" ", left=None, right=None):
pass
# morris traversal
class Solution(object):
def checkEquivalence(self, root1, root2):
"""
:type root1: Node
:type root2: Node
:rtype: bool
"""
def add_counter(counter, prev, d, val):
if val.isalpha():
counter[ord(val)-ord('a')] += d if prev[0] == '+' else -d
prev[0] = val
def morris_inorder_traversal(root, cb):
curr = root
while curr:
if curr.left is None:
cb(curr.val)
curr = curr.right
else:
node = curr.left
while node.right and node.right != curr:
node = node.right
if node.right is None:
node.right = curr
curr = curr.left
else:
cb(curr.val)
node.right = None
curr = curr.right
counter = collections.defaultdict(int)
morris_inorder_traversal(root1, functools.partial(add_counter, counter, ['+'], 1))
morris_inorder_traversal(root2, functools.partial(add_counter, counter, ['+'], -1))
return all(v == 0 for v in counter.itervalues())
# Time: O(n)
# Space: O(h)
import collections
import functools
class Solution2(object):
def checkEquivalence(self, root1, root2):
"""
:type root1: Node
:type root2: Node
:rtype: bool
"""
def add_counter(counter, prev, d, val):
if val.isalpha():
counter[ord(val)-ord('a')] += d if prev[0] == '+' else -d
prev[0] = val
def inorder_traversal(root, cb):
def traverseLeft(node, stk):
while node:
stk.append(node)
node = node.left
stk = []
traverseLeft(root, stk)
while stk:
curr = stk.pop()
cb(curr.val)
traverseLeft(curr.right, stk)
counter = collections.defaultdict(int)
inorder_traversal(root1, functools.partial(add_counter, counter, ['+'], 1))
inorder_traversal(root2, functools.partial(add_counter, counter, ['+'], -1))
return all(v == 0 for v in counter.itervalues())