LeetCode-in-Ruby.github.io
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.
# class TreeNode
# attr_accessor :val, :left, :right
# def initialize(val = 0, left = nil, right = nil)
# @val = val
# @left = left
# @right = right
# end
# end
# @param {TreeNode} root
# @return {Boolean}
def is_symmetric(root)
return true if root.nil?
helper_symmetric(root.left, root.right)
end
private
def helper_symmetric(left_node, right_node)
return left_node.nil? && right_node.nil? if left_node.nil? || right_node.nil?
return false if left_node.val != right_node.val
helper_symmetric(left_node.left, right_node.right) && helper_symmetric(left_node.right, right_node.left)
end