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symmetric-tree.py
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symmetric-tree.py
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from __future__ import print_function
# Time: O(n)
# Space: O(h), h is height of binary tree
# Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
#
# For example, this binary tree is symmetric:
#
# 1
# / \
# 2 2
# / \ / \
# 3 4 4 3
# But the following is not:
# 1
# / \
# 2 2
# \ \
# 3 3
# Note:
# Bonus points if you could solve it both recursively and iteratively.
#
# Definition for a binary tree node
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
# Iterative solution
class Solution:
# @param root, a tree node
# @return a boolean
def isSymmetric(self, root):
if root is None:
return True
stack = []
stack.append(root.left)
stack.append(root.right)
while stack:
p, q = stack.pop(), stack.pop()
if p is None and q is None:
continue
if p is None or q is None or p.val != q.val:
return False
stack.append(p.left)
stack.append(q.right)
stack.append(p.right)
stack.append(q.left)
return True
# Recursive solution
class Solution2:
# @param root, a tree node
# @return a boolean
def isSymmetric(self, root):
if root is None:
return True
return self.isSymmetricRecu(root.left, root.right)
def isSymmetricRecu(self, left, right):
if left is None and right is None:
return True
if left is None or right is None or left.val != right.val:
return False
return self.isSymmetricRecu(left.left, right.right) and self.isSymmetricRecu(left.right, right.left)
if __name__ == "__main__":
root = TreeNode(1)
root.left, root.right = TreeNode(2), TreeNode(2)
root.left.left, root.right.right = TreeNode(3), TreeNode(3)
root.left.right, root.right.left = TreeNode(4), TreeNode(4)
print(Solution().isSymmetric(root))