2493 Reverse Odd Levels Of Binary Tree

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.

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.

# 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]:
        


        def traverse(q, is_odd):
                temp = []
                while(q):
                        item = q.pop(0)
                        if item.right:
                            temp.append(item.right)
                        if item.left:
                            temp.append(item.left)
                if is_odd:
                    for idx, k in enumerate(temp):
                        temp[idx].val,temp[len(temp) - 1- idx].val = temp[len(temp) - 1- idx].val,temp[idx].val
                        if idx == (len(temp) -1) // 2:
                            break

                if temp:
                    traverse(temp, not is_odd)
        traverse([root], True)
        return root