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Description

Given the root of a perfect binary tree, reverse the node values at each odd level of the tree.

For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18], then it should become [18,29,11,7,4,3,1,2].
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.

Example 1:

Input: root = [2,3,5,8,13,21,34]
Output: [2,5,3,8,13,21,34]
Explanation: 
The tree has only one odd level.
The nodes at level 1 are 3, 5 respectively, which are reversed and become 5, 3.

Example 2:

Input: root = [7,13,11]
Output: [7,11,13]
Explanation: 
The nodes at level 1 are 13, 11, which are reversed and become 11, 13.

Example 3:

Input: root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
Output: [0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
Explanation: 
The odd levels have non-zero values.
The nodes at level 1 were 1, 2, and are 2, 1 after the reversal.
The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1 after the reversal.

Constraints:

The number of nodes in the tree is in the range [1, 2^14].
0 <= Node.val <= 105
root is a perfect binary tree.

Solution

Level traversal, use an extra space to store all the nodes in current level, and use 2 pointers starting from the sides to the middle, swapping the values.

Time complexity: $o(n)$
Space complexity: $o(\log n)$

Code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def reverseOddLevels(self, root: Optional[TreeNode]) -> Optional[TreeNode]:cur_level_nodes = []queue = collections.deque([(root, 0)])cur_level = -1while queue:node, level = queue.popleft()if level != cur_level:left, right = 0, len(cur_level_nodes) - 1while left <= right:cur_level_nodes[left].val, cur_level_nodes[right].val = cur_level_nodes[right].val, cur_level_nodes[left].valleft += 1right -= 1cur_level = levelif level & 1 == 1:cur_level_nodes.append(node)else:cur_level_nodes = []if node.left:queue.append((node.left, level + 1))queue.append((node.right, level + 1))left, right = 0, len(cur_level_nodes) - 1while left <= right:cur_level_nodes[left].val, cur_level_nodes[right].val = cur_level_nodes[right].val, cur_level_nodes[left].valleft += 1right -= 1return root