LeetCode in Elixir
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?
Solution
# Definition for a binary tree node.
#
# defmodule TreeNode do
# @type t :: %__MODULE__{
# val: integer,
# left: TreeNode.t() | nil,
# right: TreeNode.t() | nil
# }
# defstruct val: 0, left: nil, right: nil
# end
defmodule Solution do
@spec is_symmetric(root :: TreeNode.t() | nil) :: boolean
def is_symmetric(nil), do: true
def is_symmetric(%TreeNode{left: left, right: right}) do
helper(left, right)
end
defp helper(nil, nil), do: true
defp helper(nil, _), do: false
defp helper(_, nil), do: false
defp helper(%TreeNode{val: val1, left: left1, right: right1}, %TreeNode{val: val2, left: left2, right: right2}) do
val1 == val2 and helper(left1, right2) and helper(right1, left2)
end
end