solve: binaryTreeLevelOrderTraversal

Author: yolophg

binary-tree-level-order-traversal/yolophg.py

@@ -0,0 +1,36 @@
+# Time Complexity: O(n) -> visit each node exactly once.
+# Space Complexity: O(n) -> in the worst case, the queue holds all nodes at the last level.
+
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        
+        # list to store level order traversal.
+        ans = [] 
+        
+        # if the tree is empty, return an empty list.
+        if root is None:
+            return ans 
+        
+        # queue to process nodes level by level.
+        q = deque([root]) 
+        
+        while q:
+            # list to store values of nodes at the current level.
+            t = [] 
+            
+            # process all nodes at this level.
+            for _ in range(len(q)):  
+                # pop the first node from the queue.
+                node = q.popleft()  
+                # add node's value to the list.
+                t.append(node.val) 
+                
+                if node.left:
+                    q.append(node.left) 
+                if node.right:
+                    q.append(node.right) 
+            
+            # add current level values to the result list.
+            ans.append(t) 
+        
+        return ans