101. Symmetric Tree

Easy

---

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?

Recursion

# 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 isSymmetric(self, root: Optional[TreeNode]) -> bool:
        if root is None:
            return True
        def recursion(left, right):
            if left is None and right is None:
                return True
            elif left is None or right is None:
                return False
            elif left.val != right.val:
                return False
            return recursion(left.right, right.left) and recursion(left.left, right.right)
        return recursion(root.left, root.right)
            

Iterative

# 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 isSymmetric(self, root: Optional[TreeNode]) -> bool:
        # .
        if root is None:
            return True
        q = [[root.left, root.right]]
        while q:
            left, right = q.pop(0)
            if left is None and right is None:
                continue
            elif left is None or right is None:
                return False
            elif left.val != right.val:
                return False
            else:
                q.append([left.right, right.left])
                q.append([left.left, right.right])
        return True