docs(docs): add documentation for Binary Tree Level Order Traversal (LeetCode 102)

apps/docs/content/(tree)/102-binary-tree-level-order-traversal.mdx

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+---
+title: '102. Binary Tree Level Order Traversal'
+description: 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)
+sidebar:
+  label: 'Binary Tree Level Order Traversal'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Tree BFS</Badge>
+
+### 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`
+
+## Solution
+
+```py
+# 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
+```