From 315efee66ebcdd5eaa0d42c66ca802120c4596ab Mon Sep 17 00:00:00 2001
From: jeetusingh247 <jeetusingh11010@gmail.com>
Date: Sun, 8 Feb 2026 17:37:19 +0530
Subject: [PATCH] solved

---
 Binary_level_order.java | 36 +++++++++++++++++++++++++++++++
 Course_Schedule.java    | 47 +++++++++++++++++++++++++++++++++++++++++
 2 files changed, 83 insertions(+)
 create mode 100644 Binary_level_order.java
 create mode 100644 Course_Schedule.java

diff --git a/Binary_level_order.java b/Binary_level_order.java
new file mode 100644
index 00000000..282349c0
--- /dev/null
+++ b/Binary_level_order.java
@@ -0,0 +1,36 @@
+// Time : O(n) , since we are traversing each element in the tree
+// Space: O(n/2) --> O(n), since we are counting child nodes 
+// on popping each parent node going down level by level
+// Approach To Solve: solved it using breadth-first-search or bfs or level
+// order traversal using FIFO(first in first out) using queue, we can
+// also solve using DFS as well.
+
+class Solution {
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        List<List<Integer>> result = new ArrayList<>();
+        if(root == null) return result;
+
+        Queue<TreeNode> q = new LinkedList<>();
+        q.add(root);
+
+        while(!q.isEmpty()){
+            int size = q.size();
+            List<Integer> temp = new ArrayList<>();
+            for(int i=0; i<size; i++){
+                TreeNode curr = q.poll();
+                temp.add(curr.val);
+
+                if(curr.left != null){
+                    q.add(curr.left);
+                }
+
+                if(curr.right != null){
+                    q.add(curr.right);
+                }
+            }
+
+            result.add(temp);
+        }
+        return result;
+    }
+}
\ No newline at end of file
diff --git a/Course_Schedule.java b/Course_Schedule.java
new file mode 100644
index 00000000..f6535063
--- /dev/null
+++ b/Course_Schedule.java
@@ -0,0 +1,47 @@
+// Approach: solved using kahns algorithm --> BFS topological sorting
+// Time : O(V + E) where V is numCourses and E is the number of prerequisite pairs.
+// Space complexity is O(V + E).
+// The indegree array uses O(V).
+// The adjacency list stored in the HashMap uses O(E).
+// The queue can hold up to O(V) nodes in the worst case.
+// No extra recursive stack since this is BFS.
+
+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) {
+            indegrees[pr[0]]++;
+            map.putIfAbsent(pr[1], new ArrayList<>());
+            map.get(pr[1]).add(pr[0]);   // fixed syntax
+        }
+
+        int count = 0;
+        Queue<Integer> q = new LinkedList<>();
+
+        // only add nodes with indegree 0
+        for (int i = 0; i < numCourses; i++) {
+            if (indegrees[i] == 0) {
+                q.add(i);
+            }
+        }
+
+        while (!q.isEmpty()) {
+            int curr = q.poll();
+            count++;
+
+            List<Integer> dependencies = map.get(curr);
+            if (dependencies != null) {
+                for (int dependent : dependencies) {
+                    indegrees[dependent]--;
+                    if (indegrees[dependent] == 0) {
+                        q.add(dependent);
+                    }
+                }
+            }
+        }
+
+        return count == numCourses;
+    }
+}
\ No newline at end of file