diff --git a/CourseSchedule.java b/CourseSchedule.java
new file mode 100644
index 00000000..4b98a8b9
--- /dev/null
+++ b/CourseSchedule.java
@@ -0,0 +1,53 @@
+// Time Complexity : O(V+E).
+// Space Complexity : O(V+E)
+// Did this code successfully run on Leetcode : Yes
+// Approach : We first build a hashmap to represent which courses depend on which prerequisites, and an indegrees array to count how many prerequisites each course has.
+//Next, we enqueue all courses with zero indegree (no prerequisites) — these can be taken first.
+//Using BFS, we repeatedly dequeue a course, reduce the indegree of its dependent courses, and enqueue them once their indegree becomes zero.
+//If we process all numCourses this way, it means all courses can be completed and no cycle can be formed.
+
+class Solution {
+    public boolean canFinish(int numCourses, int[][] prerequisites) {
+
+        int[] indegrees = new int[numCourses];
+        HashMap<Integer, List<Integer>> map = new HashMap<>();
+
+        for(int[] pr : prerequisites){  // pr[0] is dependent courses, pr[1] is independent courses
+
+            indegrees[pr[0]]++; //construct indegrees array
+            map.putIfAbsent(pr[1], new ArrayList<>()); //construct hashmap
+            map.get(pr[1]).add(pr[0]);
+        }
+
+        int count = 0;
+        Queue<Integer> q = new LinkedList<>();
+
+        for(int i = 0; i < numCourses; i++){
+            if(indegrees[i] == 0){ //add indegrees elements to queue
+                q.add(i);
+                count++;
+            }
+        }
+
+        if(q.isEmpty()) return false;
+        if(count == numCourses) return true;
+
+        //BFS technique with queue
+        while(!q.isEmpty()){
+            int curr = q.poll();
+            List<Integer> dependencies = map.get(curr);
+            if(dependencies != null){
+                for(int dependent : dependencies){
+                    indegrees[dependent]--;
+                    if(indegrees[dependent] == 0){
+                        q.add(dependent);
+                        count++;
+                        if(count == numCourses) return true;
+                    }
+                }
+            }
+        }
+
+        return false;
+    }
+}
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diff --git a/LevelOrderTraversal.java b/LevelOrderTraversal.java
new file mode 100644
index 00000000..62d66297
--- /dev/null
+++ b/LevelOrderTraversal.java
@@ -0,0 +1,34 @@
+// Time Complexity : O(n).
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : Yes
+// Approach : I have used BFS approach to solve this with the help of queue as I can achieve FIFO. We add nodes to the queue and maintain size to determine
+// the level. And we remove node from queue, for each node we reduce the size and do this until complete level is removed. Along woth this we add the corresponding
+// left and right values to the queue as well and continue the process.
+
+class Solution {
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        List<List<Integer>> result = new ArrayList<>();
+        if(root == null){
+            return result;
+        }
+        Queue<TreeNode> queue = new LinkedList<>(); //initialize queue
+        queue.add(root); //add root to queue
+        while(!queue.isEmpty()){
+            int size = queue.size(); //maintain size for each level
+            List<Integer> temp = new ArrayList<>(); //temp List to store each level
+            while(size > 0){
+                TreeNode curr = queue.poll(); //remove one root
+                temp.add(curr.val);
+                if(curr.left != null){ //add corresponding left
+                    queue.add(curr.left);
+                }
+                if(curr.right != null){ //add corresponding right
+                    queue.add(curr.right);
+                }
+                size--; //reduce size as we poll each node
+            }
+            result.add(temp); // add the temp list to final result List
+        }
+        return result;
+    }
+}
\ No newline at end of file