Day19- Binary-tree-level-order-traversal, Course-schedule

Binary-tree-level-order-traversal.java

@@ -0,0 +1,41 @@
+/**
+ * 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 {
+    //BFS
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        List<List<Integer>> result = new ArrayList<>();
+        //Q is an interface. So it is not new Q<>()
+        Queue<TreeNode> q = new LinkedList<>();
+
+        if(root == null) return result;
+        //Q push is offer
+        q.offer(root);
+        while(!q.isEmpty()){
+            int size = q.size();
+            List<Integer> list = new ArrayList<>();
+            for(int i=0;i<size;i++){
+                TreeNode node = q.poll();
+                list.add(node.val);
+                if(node.left!=null) q.offer(node.left);
+                if(node.right!=null) q.offer(node.right);
+            }
+
+            result.add(list);
+
+        }
+        return result;
+    }
+}

Course-schedule.java

@@ -0,0 +1,50 @@
+class Solution {
+    public boolean canFinish(int numCourses, int[][] prerequisites) {
+        HashMap<Integer, List<Integer>> map = new HashMap<>();
+        int[] indegarr = new int[numCourses];
+        for(int[] prerequisite: prerequisites){
+            int indep = prerequisite[1];
+            int dep = prerequisite[0];
+            indegarr[dep]++; //in degree array will keep all inwards arrow count
+            //creating the hashmap
+            if(!map.containsKey(indep)){
+                //create new list
+                map.put(indep, new ArrayList<>());
+
+            }
+            map.get(indep).add(dep);
+        }
+
+        Queue<Integer> q = new LinkedList<>();
+
+        int count = 0;
+        for(int i = 0 ;i<numCourses; i++){
+            //if no dep add it to q
+            if(indegarr[i]== 0){
+                q.add(i);
+                count++;
+            }
+        }
+
+        if(count==numCourses) return true;
+        while(!q.isEmpty()){
+            int curr = q.poll();
+            //look for dependents in map
+            List<Integer> children = map.get(curr);
+            //reduce their indegrees by 1
+            if(children!=null){
+                for( int child: children){
+                    indegarr[child]--;
+                    if(indegarr[child]==0){
+                        q.add(child);
+                        count++;
+                    }
+                    if(count==numCourses) return true;
+                }
+            }
+
+
+        }
+        return false;
+    }
+}

README.md

@@ -1,8 +1,13 @@
 # BFS-1
 # Problem 1
 Binary Tree Level Order Traversal (https://leetcode.com/problems/binary-tree-level-order-traversal/)
+* BFS solution - add the root element in the queue first. Assign the size of queue & list for every level.
+* pop the element, add to the list & add children to the queue.
+* T- O(n) all nodes to be iterated, S - O(n) width of the tree.
 
 # Problem 2
 Course Schedule (https://leetcode.com/problems/course-schedule/)
-
+* DS1 - in degree array will keep all inwards arrow count, DS2 - HashMap(indep, list of dep), DS3 - Queue that will store V with indegrees[]==0
+* In the Q, look for dependents in map, reduce indegree by 1, Queue that will store V with indegrees[]==0
+* if count=numCourses means we can finish all courses.