Done BFS 2

Problem1.py

@@ -0,0 +1,37 @@
+#Binary Tree Right Side View
+
+# 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 -> On
+# Space complexity -> On
+# Logic -> Use levelOrderTraversal to maintaing all the elements of the same level in a same level and do it in such a way 
+        #  that always rightmost is enetered 1st then the 1st element will always be the 1 that will be seen from the right
+
+from collections import deque
+class Solution:
+    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
+
+        result = []
+        levelQueue = deque()
+        levelQueue.append(root)
+
+        while len(levelQueue) > 0:
+            itemsInCurrentLevel = len(levelQueue)
+
+            for i in range(itemsInCurrentLevel):
+                node = levelQueue.popleft()
+                if i == 0:
+                    result.append(node.val)
+                if node.right:
+                    levelQueue.append(node.right)
+
+                if node.left:
+                    levelQueue.append(node.left)
+
+        return result
+

Problem2.py

@@ -0,0 +1,46 @@
+#Cousins in Binary Tree
+
+# 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 -> On
+# Space -> On
+
+from collections import deque
+
+class Solution:
+    def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
+        # get the elements at depth A usign elvel order traversal, if both elements at that level
+        # but have different parents then they are cousins else not
+
+        levelQueue = deque()
+
+        levelQueue.append(root)
+        foundCousingX = False
+        foundCousingY = False
+        while levelQueue:
+            totalItemsInCurrentLevel = len(levelQueue)
+            for i in range(totalItemsInCurrentLevel):
+                node = levelQueue.popleft()
+
+                # check for same parent
+                if node.left and node.right:
+                    vals = {node.left.val, node.right.val}
+                    if x in vals and y in vals:
+                        return False
+                # check if found val then update the flag
+                for child in filter(None, [node.left, node.right]):
+                    foundCousingX =  child.val == x or foundCousingX
+                    foundCousingY = child.val == y or foundCousingY
+                    levelQueue.append(child)
+
+            # if in the last level found both and reached here means sibling
+            if foundCousingX and foundCousingY:
+                return True
+            # if in the last level found only 1 then not sibling
+            if foundCousingX or foundCousingY:
+                return False