Adding Simple Solution for Problem - 102 - BT Tree Level Order Traversal

Author: Vanditg

BT_Level_Order_Traversal/SimpleSolution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 102
+## Problem Name: Binary Tree Level Order Traversal 
+##===================================
+#
+#Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
+#
+#For example:
+#Given binary tree [3,9,20,null,null,15,7],
+#    3
+#   / \
+#  9  20
+#    /  \
+#   15   7
+#return its level order traversal as:
+#[
+#  [3],
+#  [9,20],
+#  [15,7]
+#]
+import queue
+class Solution:
+    def levelOrder(self, root):
+        tmp = []	#Initialize tmp empty list
+        q = queue.Queue()	#Initialize queue
+        if root is None:	#Condition-check: If root is empty
+            return None	#Return nothing 
+        q.put(root)	#Put root in our queue
+        while not q.empty():	#While queue is not empty
+            array = []	#Initialize an empty array for levelorder nodes 
+            size = q.qsize()	#Get the size of queue
+            while size != 0:	#Loop till size gets zero
+                curr = q.get()	#Initialize curr which takes the value we have put 
+                array.append(curr.val)	#Append that curr's value in array
+                if curr.left is not None:	#Condition-check: If we have left subtree or child 
+                    q.put(curr.left)	#We simply add that curr's left in queue
+                if curr.right is not None:	#Condition-check: If we have right subtree or child
+                    q.put(curr.right)	#We simply add that curr's right in queue                  
+                size -= 1	#Reduce size by 1 
+            if len(array) != 0:	#Condition-check: If the array's length is not equal to zero 
+                tmp.append(array)	#We append that array in our tmp array, which we'll return 
+        return tmp	#Finally we'll return tmp
+