BFS-1

Problem1.py

@@ -0,0 +1,63 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right 
+
+# Time Complexity --> O(n) where n is the number of nodes
+# Space Complexity --> O(n/2) 
+# Approach --> BFS
+from collections import deque
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        if root is None:
+            return []
+        q = deque()
+        q.append(root)
+        result = []
+        while q:
+            size = len(q)
+            temp = []
+            for i in range(size):
+                curr = q.popleft()
+                temp.append(curr.val)
+                if curr.left is not None:
+                    q.append(curr.left)
+                if curr.right is not None:
+                    q.append(curr.right) 
+            result.append(temp)
+        return result 
+
+'''
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
+# Time Complexity --> O(n) where n is the number of nodes
+# Space Complexity --> O(logn) which is the recursive stack size(height of the tree)
+# Approach --> DFS
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        self.result = []
+        self.helper(root, 0)
+        return self.result 
+
+    def helper(self, root, level):
+        # base
+        if root is None:
+            return
+
+        # logic
+        if len(self.result)==level:
+            self.result.append([])
+        self.result[level].append(root.val)
+
+        self.helper(root.left, level+1)
+        self.helper(root.right, level+1)
+
+
+'''

Problem2.py

@@ -0,0 +1,39 @@
+# Time Complexity --> O(V + E) where v is the number of vertices or courses and E is the number of edges or dependencies
+# Space Complexity --> O(V + E)
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+
+        hmap = {}
+        indegree = [0 for i in range(numCourses)]
+        for prereq in prerequisites:
+            if prereq[1] not in hmap:
+                hmap[prereq[1]] = []
+            hmap[prereq[1]].append(prereq[0]) # Independent(Prereq) to Dependent mapping
+            indegree[prereq[0]] += 1 # keep track of indegree
+
+        q = deque()
+        count = 0
+        for i in range(len(indegree)):
+            if indegree[i]==0:
+                q.append(i)
+                count += 1
+
+        if count==numCourses: # If there are no courses with prereqs then return true
+            return True 
+        if len(q)==0:   # If there are no courses with 0 prereqs, then it is a cyclic graph and return False
+            return False
+
+        # Start with the independent courses and then look over all their dependent courses, reduce each of indegree by 1 and if all the indegrees get to 0 then return True
+        while q:
+            curr = q.popleft()
+            if curr in hmap and len(hmap[curr])>0:
+                for course in hmap[curr]:
+                    indegree[course] -= 1
+                    if indegree[course]==0:
+                        q.append(course)
+                        count += 1
+                    if count==numCourses:
+                        return True
+        return False 
+
+