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# 2415. Reverse Odd Levels of Binary Tree
"""
given the 'root' of a perfect binary tree, reverse the node values at each
odd level of the tree. return the root of the reversed tree. a binary tree is
perfect if all parent nodes have two children and all leaves are on the same
level. the level of a node is the number of edges along the path between it
and the root node.
"""
# Definition for a binary tree node.
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def reverseOddLevels(self, root):
"""
:type root: Optional[TreeNode]
:rtype: Optional[TreeNode]
"""
def dfs(node1, node2, level):
if not node1 or not node2:
return
if level % 2 != 0:
node1.val, node2.val = node2.val, node1.val
dfs(node1.left, node2.right, level + 1)
dfs(node1.right, node2.left, level + 1)
dfs(root.left, root.right, 1)
return root
if __name__ == "__main__":
obj = Solution()
print(obj.reverseOddLevels(root=[2, 3, 5, 8, 13, 21, 34]))
print(obj.reverseOddLevels(root=[7, 13, 11]))
print(obj.reverseOddLevels(root=[0, 1, 2, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2]))
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