Problem Statement
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1 / \ 2 2 \ \ 3 3
Follow up: Solve it both recursively and iteratively.
Solution:
from collections import deque
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
#Recursive approach
class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
return self.check(root.left, root.right)
def check(self, left, right):
if not left and not right:
return True
if not left or not right:
return False
if left.val == right.val:
valid1 = self.check(left.left, right.right)
valid2 = self.check(left.right, right.left)
return valid1 and valid2
return False
#Iterative apporach
class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
q = deque()
q.append(root)
q.append(root)
while len(q)>0:
n1 = q.popleft()
n2 = q.popleft()
if not n1 and not n2:
continue
if not n1 or not n2:
return False
if n1.val!=n2.val:
return False
q.append(n1.left)
q.append(n2.right)
q.append(n1.right)
q.append(n2.left)
return True
