Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3]
Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3]
Output: false
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. -
-100 <= Node.val <= 100
Follow up: Could you solve it both recursively and iteratively?SOLUTION:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isMirror(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
if not p or not q:
if p or q:
return False
return True
if p.val == q.val and self.isMirror(p.left, q.right) and self.isMirror(p.right, q.left):
return True
def isSymmetric(self, root: Optional[TreeNode]) -> bool:
if root:
return self.isMirror(root.left, root.right)
return True
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