diff --git a/bt-level-order.py b/bt-level-order.py
new file mode 100644
index 00000000..df7db410
--- /dev/null
+++ b/bt-level-order.py
@@ -0,0 +1,30 @@
+# 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
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+         # bfs - use queue fifo, popleft()
+        # when queue is empty, we know level is complete
+        levels=[]
+        if not root:
+            return levels
+        
+        # levels = [[3],[]]
+        # q = [9,20]
+        level=0
+        queue = deque([root])
+        while queue:
+            levels.append([])
+            for i in range(len(queue)):
+                node = queue.popleft()
+                levels[level].append(node.val)
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+            level+=1
+
+        return levels
\ No newline at end of file
diff --git a/course-sched.py b/course-sched.py
new file mode 100644
index 00000000..298a4273
--- /dev/null
+++ b/course-sched.py
@@ -0,0 +1,90 @@
+# Keep track of indegrees using topological sort
+
+# BFS
+# O(V+E) time, O(V+E) space
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        indegrees = [0] * numCourses
+        graph = {}
+
+        for pr in prerequisites:
+            # build the topological sort
+            indegrees[pr[0]] += 1
+            # build the adjacency list
+            if pr[1] not in graph:
+                graph[pr[1]] = []
+            graph[pr[1]].append(pr[0])
+
+        count = 0
+        q = deque()
+
+        for i in range(numCourses):
+            # finds the roots - which have 0 prerequisites. Side note - in this problem 1 graph can have many roots
+            if indegrees[i] == 0:
+                q.append(i)
+                count += 1
+
+        # every single course has at least 1 prerequisite (deadlock cycle)
+        if not q:
+            return False
+        # all courses have 0 prereqs. Every course is already ready to take - no need for bfs 
+        if count == numCourses:
+            return True
+
+        while q:
+            curr = q.popleft()
+            dependencies = graph.get(curr)
+            if dependencies:
+                for dep in dependencies:
+                    indegrees[dep] -= 1
+                    if indegrees[dep] == 0:
+                        q.append(dep)
+                        count += 1
+                        # to finish all courses, count must eventually equal numCourses. 
+                        # as soon as you decrement indegree to 0 for a course, you add it to the queue and increment count
+                        # if count == numCourses, you already found a valid path for all courses 
+                        if count == numCourses:
+                            return True
+
+        return False
+
+
+# DFS O(V+E) time, O(V+E) space
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        self.map = {}
+        self.path = [False] * numCourses
+        self.visited = [False] * numCourses
+
+        for pr in prerequisites:
+            if pr[1] not in self.map:
+                self.map[pr[1]] = []
+            self.map[pr[1]].append(pr[0])
+
+        for i in range(numCourses):
+            if self.hasCycle(i):
+                return False
+
+        return True
+
+    def hasCycle(self, i: int) -> bool:
+        if self.visited[i]:
+            return False
+
+        if self.path[i]:
+            return True
+
+        self.path[i] = True
+
+        neighbours = self.map.get(i)
+        if neighbours is not None:
+            for ne in neighbours:
+                if self.hasCycle(ne):
+                    return True
+
+        # you could have A->B->C and A->C. if you don't set path c back to false after exploring through b, the algorithm would think a-c is a cycle 
+        self.path[i] = False
+        self.visited[i] = True
+
+        return False
+
diff --git a/right-side-view.py b/right-side-view.py
new file mode 100644
index 00000000..22e9ce1f
--- /dev/null
+++ b/right-side-view.py
@@ -0,0 +1,45 @@
+# O(n) time, O(h) space
+# DFS
+# 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
+class Solution:
+    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
+        result = []
+        self.helper(root,0,result)
+        return result 
+
+    def helper(self, root, level, result):
+        if not root:
+            return 
+
+        if len(result)==level:
+            result.append(root.val)
+
+        self.helper(root.right,level+1,result)
+        self.helper(root.left, level+1, result)
+
+# BFS
+class Solution:
+    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
+        res = []
+        q = deque([root])
+
+        while q:
+            rightSide = None 
+            for i in range(len(q)):
+                node = q.popleft()
+                if node:
+                    rightSide=node
+                    if node.left:
+                        q.append(node.left)
+                    if node.right:
+                        q.append(node.right)
+
+            if rightSide: 
+                res.append(rightSide.val)
+
+        return res