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Solution.rb
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80 lines (53 loc) · 1.05 KB
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# Definition for a binary tree node.
class TreeNode
attr_accessor :val, :left, :right
def initialize(val)
@val = val
@left, @right = nil, nil
end
end
# @param {TreeNode} root
# @return {Boolean}
def is_symmetric(root)
if root.nil?
return true
end
if root.left.nil? and root.right.nil?
return true
end
leftVector = []
rightVector = []
collect_left root.left, leftVector
collect_right root.right, rightVector
if leftVector.size != rightVector.size
return false
end
index = 0
bound = leftVector.size
while index < bound and leftVector[index] == rightVector[index]
index += 1
end
if index != bound
false
else
true
end
end
def collect_left(node, vector)
if node.nil?
vector << '#'
return
end
vector << node.val
collect_left node.left, vector
collect_left node.right, vector
end
def collect_right(node, vector)
if node.nil?
vector << '#'
return
end
vector << node.val
collect_right node.right, vector
collect_right node.left, vector
end