medium tree, graph, dp questions

backend/data/questions/binary-tree-level-order.yaml

@@ -0,0 +1,135 @@
+title: Binary Tree Level Order Traversal
+slug: binary-tree-level-order
+difficulty: medium
+leetcode_id: 102
+leetcode_url: https://leetcode.com/problems/binary-tree-level-order-traversal/
+categories:
+  - trees
+  - queue
+patterns:
+  - bfs
+
+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).
+
+constraints: |
+  - The number of nodes in the tree is in the range [0, 2000].
+  - -1000 <= Node.val <= 1000
+
+examples:
+  - input: "root = [3,9,20,null,null,15,7]"
+    output: "[[3],[9,20],[15,7]]"
+    explanation: "Level 0 has 3, level 1 has 9 and 20, level 2 has 15 and 7."
+  - input: "root = [1]"
+    output: "[[1]]"
+    explanation: "Single node at level 0."
+  - input: "root = []"
+    output: "[]"
+    explanation: "Empty tree returns empty list."
+
+explanation:
+  approach: |
+    1. Use a queue for BFS traversal
+    2. Track the number of nodes at current level
+    3. Process all nodes at current level before moving to next
+    4. Add children to queue as we process each node
+    5. Collect values for each level in a separate list
+
+  intuition: |
+    BFS naturally visits nodes level by level. By tracking how many nodes are in the queue
+    at the start of each level, we know exactly when one level ends and the next begins.
+
+    The key insight is that after processing all nodes of level k, the queue contains
+    exactly all nodes of level k+1.
+
+  common_pitfalls:
+    - title: Not tracking level boundaries
+      description: |
+        Without tracking level size, you can't separate nodes into their levels.
+        Capture queue size at the start of each level iteration.
+      wrong_approach: "Processing queue without counting level size"
+      correct_approach: "level_size = len(queue); process level_size nodes"
+
+    - title: Forgetting null check
+      description: |
+        Empty tree (null root) should return empty list, not cause an error.
+
+    - title: Using list as queue
+      description: |
+        Using list.pop(0) is O(n). Use collections.deque for O(1) popleft.
+
+  key_takeaways:
+    - BFS with queue is the standard for level-order traversal
+    - Track level size to group nodes by level
+    - This pattern extends to many tree problems (zigzag, right side view, etc.)
+    - deque is more efficient than list for queue operations
+
+  time_complexity: "O(n)"
+  space_complexity: "O(n)"
+  complexity_explanation: |
+    Time: Visit each node exactly once.
+    Space: Queue holds at most one level of nodes, which is O(n) in worst case (complete tree).
+
+solutions:
+  - approach_name: BFS with Queue (Optimal)
+    is_optimal: true
+    code: |
+      from collections import deque
+
+      class TreeNode:
+          def __init__(self, val=0, left=None, right=None):
+              self.val = val
+              self.left = left
+              self.right = right
+
+      def level_order(root: TreeNode | None) -> list[list[int]]:
+          if not root:
+              return []
+
+          result = []
+          queue = deque([root])
+
+          while queue:
+              level_size = len(queue)
+              level_values = []
+
+              for _ in range(level_size):
+                  node = queue.popleft()
+                  level_values.append(node.val)
+
+                  if node.left:
+                      queue.append(node.left)
+                  if node.right:
+                      queue.append(node.right)
+
+              result.append(level_values)
+
+          return result
+    explanation: |
+      Process nodes level by level using a queue.
+      Track level size to know when to start a new level list.
+
+  - approach_name: DFS with Level Tracking
+    is_optimal: false
+    code: |
+      def level_order(root: TreeNode | None) -> list[list[int]]:
+          result = []
+
+          def dfs(node: TreeNode | None, level: int) -> None:
+              if not node:
+                  return
+
+              if level == len(result):
+                  result.append([])
+
+              result[level].append(node.val)
+
+              dfs(node.left, level + 1)
+              dfs(node.right, level + 1)
+
+          dfs(root, 0)
+          return result
+    explanation: |
+      DFS alternative: pass level as parameter and append to appropriate list.
+      Same time complexity but uses recursion stack space.