diff --git a/BinaryRightSide.py b/BinaryRightSide.py
new file mode 100644
index 00000000..c199a5f7
--- /dev/null
+++ b/BinaryRightSide.py
@@ -0,0 +1,74 @@
+
+from typing import List, Optional
+from collections import deque
+
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+class BFSSolution:
+# Time Complexity : O(n), where n is the number of nodes in the binary tree
+# Space Complexity : O(w), where w is the maximum width of the binary tree
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Your code here along with comments explaining your approach
+
+    # BFS Approach
+    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
+        q = deque() # Init queue for level order traversal
+        result = [] # init result list
+
+        if root is None: # Handling edge case
+            return result
+        # Adding the root element to queue
+        q.append(root)
+
+        # Iterating level by level
+        while len(q):
+            size = len(q) # Getting the size for traversing the partciular level
+            for i in range(size): # Iterating the level
+                curr = q.popleft() # Popping the left most element
+                if i == size-1: # if last element, then store it in the list
+                    result.append(curr.val)
+                if curr.left is not None:
+                    q.append(curr.left)
+                if curr.right is not None:
+                    q.append(curr.right)
+
+        return result
+    
+class DFSSolution:
+# Time Complexity : O(n), where n is the number of nodes in the binary tree
+# Space Complexity : O(H), where H is the height of the binary tree
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Your code here along with comments explaining your approach
+
+    # DFS
+    def __init__(self):
+        self.result = []# intializing the result list in global scope because in dfs, result will be added here
+
+    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
+        self.dfs(root, 0)
+        return self.result
+    
+    def dfs(self, root: Optional[TreeNode], depth: int)-> None:
+        # Doing a In order traversal Root-->Left-->Right
+        # So at each depth, we will overwrite the previous element at same depth
+
+        # Checking null/base case
+        if root is None:
+            return
+        
+        # Checking if depth is equal to result list size, if so then add the curr element because this is first time we are reaching this depth
+        if depth == len(self.result):
+            self.result.append(root.val)
+        else: # Otherwise just overwrite the value
+            self.result[depth] = root.val
+
+        # Moving left
+        self.dfs(root.left, depth+1)
+        #Moving Right
+        self.dfs(root.right, depth+1)
\ No newline at end of file
diff --git a/CousinsBinary.py b/CousinsBinary.py
new file mode 100644
index 00000000..e1416795
--- /dev/null
+++ b/CousinsBinary.py
@@ -0,0 +1,155 @@
+from typing import Optional, List
+from collections import deque
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+class BFSSolution:
+# Time Complexity : O(n), where n is the number of nodes in the binary tree
+# Space Complexity : O(2w), where w is the maximum width of the binary tree
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Your code here along with comments explaining your approach
+
+    # BFS approach
+    def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
+        q = deque() # to store nodes while traversing level order
+        parent_q = deque() # to store parent of each node 
+        x_found = False
+        y_found = False
+        x_parent = None
+        y_parent = None
+
+        # add root to queue
+        if root is not None:
+            q.append(root)
+        # add root's parent to parentQ
+        parent_q.append(None)
+
+        while len(q):
+            # Iterate at the particular level
+            size = len(q)
+            for _ in range(size):
+                curr = q.popleft() # get current element 
+                parent = parent_q.popleft() # Get current element's parent 
+
+                # check if curr is equal to x or y
+                if curr.val == x:
+                    x_found = True
+                    x_parent = parent
+                    continue
+                elif curr.val == y:
+                    y_found = True
+                    y_parent = parent
+                    continue
+
+                if curr.left is not None: # Append left child
+                    q.append(curr.left)
+                    parent_q.append(curr)
+                
+                if curr.right is not None: # appending right child
+                    q.append(curr.right)
+                    parent_q.append(curr)
+                 
+            if x_found and y_found and (x_parent != y_parent): # CHecking at each level if x and y found and their parent not same
+                return True
+            if x_found or y_found: # If either x or y is found, but not both then fail
+                return False
+
+        return False
+    
+class BFSSolution2:
+# Time Complexity : O(n), where n is the number of nodes in the binary tree
+# Space Complexity : O(w), where w is the maximum width of the binary tree
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Your code here along with comments explaining your approach
+
+    # BFS approach
+    from collections import deque
+    def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
+        q = deque() # to store nodes while traversing level order
+        x_found = False
+        y_found = False
+
+        # add root to queue
+        if root is not None:
+            q.append(root)
+
+        while len(q):
+            # Iterate at the particular level
+            size = len(q)
+            for _ in range(size):
+                curr = q.popleft() # get current element 
+
+                # check if curr is equal to x or y
+                if curr.val == x:
+                    x_found = True
+                    continue
+                elif curr.val == y:
+                    y_found = True
+                    continue   
+
+                if curr.left is not None: # Append left child
+                    q.append(curr.left)
+                
+                if curr.right is not None: # appending right child
+                    q.append(curr.right)
+                
+                if curr.left is not None and curr.right is not None: # Eliminating the scenario where x and y are siblings, so they can't be cousins
+                    if curr.left.val == x and curr.right.val == y:
+                        return False
+                    if curr.left.val == y and curr.right.val == x:
+                        return False
+                 
+            if x_found and y_found: # CHecking at each level if x and y found and their parent not same
+                return True
+            if x_found or y_found: # If either x or y is found, but not both then fail
+                return False
+
+        return False
+    
+class DFSSolution:
+# Time Complexity : O(n), where n is the number of nodes in the binary tree
+# Space Complexity : O(H), where H is the height of the binary tree
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Your code here along with comments explaining your approach
+
+    # DFS Approach
+    def __init__(self):
+         # Initializing variables in global scope which are required in DFS for finding cousins
+        self.x_found = False
+        self.y_found = False
+        self.x_depth = -1
+        self.y_depth = -1
+        self.x_parent = None
+        self.y_parent = None
+
+    def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
+        self.dfs(root, 0, None,x , y) # Starting recursion
+
+        return self.x_found and self.y_found and (self.x_parent != self.y_parent) and (self.x_depth == self.y_depth)
+
+    def dfs(self, root: Optional[TreeNode], depth: int, parent: Optional[TreeNode], x: int, y: int)-> None:
+        # Bae case
+        if root is None:
+            return
+        if self.x_found and self.y_found: # If found, then return instead of continuing with dfs
+            return
+
+        # check if root.val is equal to x or y
+        if root.val == x:
+            self.x_found = True
+            self.x_depth = depth
+            self.x_parent = parent
+        elif root.val == y:
+            self.y_found = True
+            self.y_depth = depth
+            self.y_parent = parent
+        
+        # Move Left
+        self.dfs(root.left, depth+1, root,x,y)
+        # Move Right
+        self.dfs(root.right, depth+1, root,x,y)
\ No newline at end of file