Done BFS-1

problem1.java

@@ -0,0 +1,93 @@
+
+ /**
+    Time Complexity : O(N)
+    Explanation:
+    Each node in the tree is visited exactly once.
+
+    Space Complexity : O(N)
+    Explanation:
+    In the worst case, the recursion stack or queue can hold up to N nodes
+    (for skewed trees or wide levels).
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially implemented the BFS approach using a queue.
+    Later implemented a DFS solution by tracking the level of each node.
+    Needed to check if the current level already exists in the result list;
+    if not, create a new list for that level before adding the node value.
+    */
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+
+class Solution {
+
+
+
+    public List<List<Integer>> levelOrder(TreeNode root) {
+
+        List<List<Integer>> result = new ArrayList<>();
+        dfs(root, 0, result);
+        return result;
+    }
+
+    // BFS approach (optional alternative)
+    private List<List<Integer>> bfs(TreeNode root) {
+
+        Queue<TreeNode> q = new LinkedList<>();
+        List<List<Integer>> li = new ArrayList<>();
+
+        if (root == null) return li;
+
+        q.add(root);
+
+        while (!q.isEmpty()) {
+
+            int size = q.size();
+            List<Integer> temp = new ArrayList<>();
+
+            for (int i = 0; i < size; i++) {
+
+                TreeNode t = q.poll();
+                temp.add(t.val);
+
+                if (t.left != null) q.add(t.left);
+                if (t.right != null) q.add(t.right);
+            }
+
+            li.add(temp);
+        }
+
+        return li;
+    }
+
+    // DFS approach used in solution
+    private void dfs(TreeNode root, int level, List<List<Integer>> result) {
+
+        if (root == null) return;
+
+        // Create new level list if needed
+        if (level == result.size()) {
+            result.add(new ArrayList<>());
+        }
+
+        result.get(level).add(root.val);
+
+        dfs(root.left, level + 1, result);
+        dfs(root.right, level + 1, result);
+    }
+}

problem2.java

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+    /**
+    Time Complexity : O(V + E)
+    Explanation:
+    V = number of courses
+    E = number of prerequisite relations
+    We build the graph once and process each node and edge once
+    using Kahn’s Algorithm (Topological Sort with BFS).
+
+    Space Complexity : O(V + E)
+    Explanation:
+    - HashMap stores adjacency list of the graph
+    - Queue stores nodes with 0 indegree
+    - indegree array stores dependency count
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially confused about how to represent course dependencies.
+    Later implemented it as a directed graph where:
+        prerequisite → dependent course
+    Then applied Kahn’s Algorithm using indegree tracking.
+    Also needed a counter to check if all courses are processed,
+    which confirms there is no cycle.
+    */
+
+    class Solution {
+
+
+
+        public boolean canFinish(int numCourses, int[][] prerequisites) {
+
+            Queue<Integer> q = new LinkedList<>();
+            int n = numCourses;
+
+            int[] inDeg = new int[n];
+            HashMap<Integer, List<Integer>> map = new HashMap<>();
+
+            int cnt = 0;
+
+            // Build graph and indegree array
+            for (int[] edge : prerequisites) {
+
+                int dep = edge[0];
+                int ind = edge[1];
+
+                if (!map.containsKey(ind)) {
+                    map.put(ind, new ArrayList<>());
+                }
+
+                map.get(ind).add(dep);
+                inDeg[dep]++;
+            }
+
+            // Add nodes with indegree 0 to queue
+            for (int i = 0; i < n; i++) {
+                if (inDeg[i] == 0) {
+                    q.add(i);
+                    cnt++;
+                }
+            }
+
+            if (cnt == n) return true;
+            if (map.isEmpty()) return true;
+
+            // BFS for topological sort
+            while (!q.isEmpty()) {
+
+                int size = q.size();
+
+                for (int i = 0; i < size; i++) {
+
+                    int curr = q.poll();
+                    List<Integer> depli = map.get(curr);
+
+                    if (depli != null) {
+
+                        for (int k : depli) {
+
+                            inDeg[k]--;
+
+                            if (inDeg[k] == 0) {
+                                q.add(k);
+                                cnt++;
+
+                                if (cnt == n) return true;
+                            }
+                        }
+                    }
+                }
+            }
+
+            return false;
+        }
+    }