From 6925c6815f6fe48b39a0538cdf03cef3bb6c7805 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Tue, 26 Apr 2022 00:21:37 +0530
Subject: [PATCH] btree-level-order-traversal py3

---
 .../README.md                                 | 28 +++++++++++
 .../python3/solution.py                       | 49 +++++++++++++++++++
 2 files changed, 77 insertions(+)
 create mode 100644 0102_binary-tree-level-order-traversal/README.md
 create mode 100644 0102_binary-tree-level-order-traversal/python3/solution.py

diff --git a/0102_binary-tree-level-order-traversal/README.md b/0102_binary-tree-level-order-traversal/README.md
new file mode 100644
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@@ -0,0 +1,28 @@
+Given the `root` of a binary tree, return _the level order traversal of its nodes' values_. (i.e., from left to right, level by level).
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg)
+
+    Input: root = [3,9,20,null,null,15,7]
+    Output: [[3],[9,20],[15,7]]
+    
+
+**Example 2:**
+
+    Input: root = [1]
+    Output: [[1]]
+    
+
+**Example 3:**
+
+    Input: root = []
+    Output: []
+    
+
+**Constraints:**
+
+*   The number of nodes in the tree is in the range `[0, 2000]`.
+*   `-1000 <= Node.val <= 1000`
+
+https://leetcode.com/problems/binary-tree-level-order-traversal/
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diff --git a/0102_binary-tree-level-order-traversal/python3/solution.py b/0102_binary-tree-level-order-traversal/python3/solution.py
new file mode 100644
index 0000000..6efe0fe
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+++ b/0102_binary-tree-level-order-traversal/python3/solution.py
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+# 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
+
+from collections import deque
+
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        '''
+        BFS
+        '''
+        
+        if root is None: return []
+        
+        q = deque([root])
+        results = []
+        
+        while q:
+            n = len(q)
+            
+            level = []
+            # Idea here is we iterate through the queue, the queue
+            # will always contain 1, 2, 4, 8 ... elements for each
+            # level since we'll be appending None .left/.right nodes
+            # also. This makes it easier to do the computation.
+            for i in range(n):
+                curr = q.popleft()
+                
+                # Could be None, ignore if that's the case
+                if curr:
+                    level.append(curr.val)
+                    
+                    # It's okay to append None values here, the condition
+                    # above will filter it out
+                    q.append(curr.left)
+                    q.append(curr.right)
+            
+            # Since there's a chance that at the last level of the tree, we
+            # might be pushing None nodes due to .left/.right being None, we
+            # might have an empty level. We check and avoid adding the empty
+            # level in this case.
+            if level:
+                results.append(level)
+        
+        return results
+            
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