BFS-1 Done

binary-tree-level-order-traversal.py

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+#----------------Solution 1 : BFS level traversal-----------
+''' Time Complexity : O(n) 
+    Space Complexity : O(n) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Approach : Maintain the size of queue. After traversing one level add the list to result
+'''
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        if root is None:
+            return []
+        result = []
+        q = []
+        q.append(root)
+        while q:
+            size = len(q)
+            l = []
+            for i in range(size):
+                node = q.pop(0)
+                l.append(node.val)
+                if node.left:
+                    q.append(node.left)
+                if node.right:
+                    q.append(node.right)
+            result.append(l)
+
+        return result
+
+#----------------Solution 1 : DFS traversal-----------
+''' Time Complexity : O(n) 
+    Space Complexity : O(h) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Approach : Maintain global result, when new level is encountered create new list at the index of level.
+              Traverse the tree in preorder and append the nodes at respective index list.
+'''
+class Solution:
+    def __init__(self):
+        self.result = []
+
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+
+        def dfs(root,level):
+            if root is None:
+                return 
+            if len(self.result) == level:
+                self.result.append([])
+
+            self.result[level].append(root.val)
+            dfs(root.left,level+1)
+            dfs(root.right,level+1)
+
+        dfs(root, 0)
+        return self.result

course-schedule.py

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+#----------------Solution 1 : BFS -----------
+''' Time Complexity : O(V + E) 
+    Space Complexity : O(V + E) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Approach : Using topological sort algorithm
+'''
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        hashmap = defaultdict(list)
+        count = 0
+        indegree = [0 for _ in range(numCourses)]
+        for crs,pre in prerequisites:
+            hashmap[pre].append(crs)
+            indegree[crs] += 1 
+        q = []
+        for i in range(numCourses):
+            if indegree[i] == 0:
+                q.append(i)
+                count += 1
+        while q:
+            crs = q.pop(0)
+            for pre in hashmap[crs]:
+                indegree[pre] -= 1
+                if indegree[pre] == 0:
+                    q.append(pre)
+                    count += 1
+
+        if count == numCourses:
+            return True
+        else:
+            return False
+
+#----------------Solution 2 : DFS -----------
+''' Time Complexity : O(V + E) 
+    Space Complexity : O(V + E) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Approach : Maintaining visited set to detect cycle, Travresing each node by dfs
+'''
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        hashmap = {i:[] for i in range(numCourses)}
+        for crs,pre in prerequisites:
+            hashmap[pre].append(crs)
+
+        visited = set()
+        def dfs(crs):
+            #if no prerequisite return True
+            if hashmap[crs] == []:
+                return True
+            if crs in visited:
+                return False
+            visited.add(crs)
+            for pre in hashmap[crs]:
+                if not dfs(pre):
+                    return False
+            #if True for all prereq remove from visited
+            visited.remove(crs)
+            # Dont traverse completed crs again
+            hashmap[crs] = []
+            return True
+        #call dfs for all courses
+        for i in range(numCourses):
+            if not dfs(i):
+                return False
+        return True