# 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]))