btree-level-order-traversal py3

0102_binary-tree-level-order-traversal/python3/solution.py

@@ -0,0 +1,49 @@
+# 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
+