feat(work): add solution for Binary Tree Level Order Traversal (LC 102)

work/1/Medium/Tree/102.binary-tree-level-order-traversal.py

@@ -0,0 +1,66 @@
+"""
+102. Binary Tree Level Order Traversal
+Difficulty: Medium
+https://leetcode.com/problems/binary-tree-level-order-traversal/
+
+──────────────────────────────────────────────────
+
+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:
+
+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
+"""
+
+# 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:
+    # T: O(n) S: O(n)
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        res = []
+
+        q = deque()
+        q.append(root)
+
+        while q:
+            qLen = len(q)
+            level = []
+            for i in range(qLen):
+                node = q.popleft()
+                if node:
+                    level.append(node.val)
+                    q.append(node.left)
+                    q.append(node.right)
+            if level:
+                res.append(level)
+
+        return res