Skip to content

2415. Reverse Odd Levels of Binary Tree

Difficulty Topics

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, 214].
  • 0 <= Node.val <= 105
  • root is a perfect binary tree.

Solution

reverse-odd-levels-of-binary-tree.py
# 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]:
        dq = deque([root])
        level = 0

        while dq:
            n = len(dq)
            currLevel = []

            for _ in range(n):
                node = dq.popleft()
                currLevel.append(node)

                for child in filter(None, (node.left, node.right)):
                    dq.append(child)

            if level % 2 == 1:
                vals = [node.val for node in currLevel][::-1]
                for node, val in zip(currLevel, vals):
                    node.val = val

            level += 1

        return root