2415 Reverse Odd Levels of Binary Tree

Intuition

We are required to reverse the values on each odd level.

Let’s break this down - we need access of elements in every level? Also, we need access to the height of every level to determine if it’s odd.

We can do a level-order traversal a.k.a breadth-first search and keep track of every level. Then reverse the elements and change value on the nodes. It’s very similar to how one would reverse an array - by iterating till mid element and swapping values on left and right pointers.

Note: for Python devs, assume you are in a for loop and i is the iterator variable - ~i accesses the corresponding element from the opposite end.

Code

Python3

def reverseOddLevels(self, root: Optional[TreeNode], h: int = 0) -> Optional[TreeNode]:
 
    '''
    level order traversal
 
    1. collect all values at a level in a list
    2. reverse it and assign to the nodes
    '''
 
    queue = collections.deque([root])
 
    depth = 0
    while queue:
        n = len(queue)
 
        for _ in range(n):
            node = queue.pop()
 
            if node.left:
                queue.appendleft(node.left)
            if node.right:
                queue.appendleft(node.right )
        
        depth += 1
 
        if depth % 2 != 0:   # if level height is odd
            for i in range(len(queue)//2):
                queue[i].val, queue[~i].val = queue[~i].val, queue[i].val
    
 
    return root

Big O Analysis

• Runtime

  The runtime complexity here is $O(N)$ since we are accessing each node once in a context. We are also reversing, but that’s not nested.

• Memory

  The memory usage is $O(N)$ since we use a collections.deque (a double ended queue) to store our nodes.

— A

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