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# Level order traversal of binary tree

Given the root of a binary tree, display the node values at each level. Node values for all levels should be displayed on separate lines. Let’s take a look at the below binary tree.

Level order traversal for this tree should look like:

``````100
50, 200
25, 75, 350
``````

Runtime complexity: Linear, O(n)

Memory Complexity: Linear, O(n)

Example:

``````Input: Tree([100, [50, 25, 75], [200, None, 350]])
Output: [[100], [50, 200], [25, 75, 350]]

Input: Tree1([10, [5, [3, 2, 4], 7], [20, 15, 35]])
Output: [[10], [5, 20], [3, 7, 15, 35], [2, 4]]
``````

Constraints:

• 1<= number of nodes of the Tree <= 1000
INSTRUCTIONS

1. Write your code inside a function named `level_order_traversal`
2. There are no partial marks for the question.
3. Your function must return the output, it should not print the output.
4. To execute a block on the right-side coding panel, please press 'shift'+ 'enter' inside the block.
5. Your code should work for all permitted possible structures of the Tree.

You need to create a function with the following signature that should return an array of array. The first elements of the array would be an array containing only the top element and the last element of this array would be the array containing leaf nodes:

``````def level_order_traversal(root):
result = ""
#TODO: Write - Your - Code
return result
``````

Please create the Tree using the following code:

``````class Tree:
def __str__(self):
if self.left or self.right:
return "[%s, %s, %s]" % (self.value, self.left, self.right)
else:
return "%s" % (self.value)
def __repr__(self):
return str(self)
def __init__(self, arr):
self.left = None
self.right = None
if type(arr) is list:
self.value = arr[0]
if arr[1]:
self.left = Tree(arr[1])
if arr[2]:
self.right = Tree(arr[2])
else:
self.value = arr
``````

To create the tree shown in the below image, call:

``````Tree([100, [50, 25, 75], [200, None, 350]])
``````

No hints are availble for this assesment

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