From 966397c9618388c346bd6ba0ae54f77571380c8f Mon Sep 17 00:00:00 2001
From: Shinjanee Gupta <shinjaneegupta@gmail.com>
Date: Wed, 4 Mar 2026 18:08:18 -0800
Subject: [PATCH] BFS-2

---
 Cousins.py   | 50 ++++++++++++++++++++++++++++++++++++++++++++++++++
 RightView.py | 39 +++++++++++++++++++++++++++++++++++++++
 2 files changed, 89 insertions(+)
 create mode 100644 Cousins.py
 create mode 100644 RightView.py

diff --git a/Cousins.py b/Cousins.py
new file mode 100644
index 00000000..5865a974
--- /dev/null
+++ b/Cousins.py
@@ -0,0 +1,50 @@
+# 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 : Do a level order traversal using a queue, checking each level for x and y.
+# If they are siblings (same parent), return false immediately.
+# If both found on the same level but not siblings, return true, otherwise false.
+
+# 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
+class Solution:
+    def isCousins(self, root: Optional[TreeNode], x: int, y: int) -> bool:
+        q = deque()
+        q.append(root)
+
+        x_found, y_found = False, False
+
+        while q:
+            size = len(q)
+
+            for _ in range(size):
+                curr = q.popleft()
+
+                if curr.left and curr.right:
+                    # siblings - same parent
+                    if curr.left.val == x and curr.right.val == y:
+                        return False
+                    if curr.right.val == x and curr.left.val == y:
+                        return False
+
+                if curr.val == x:
+                    x_found = True
+                if curr.val == y:
+                    y_found = True
+
+                if curr.left:
+                    q.append(curr.left)
+                if curr.right:
+                    q.append(curr.right)
+
+            if x_found and y_found:
+                return True # same leval
+            if x_found or y_found:
+                return False # different level - not cousins
+
+        return True
\ No newline at end of file
diff --git a/RightView.py b/RightView.py
new file mode 100644
index 00000000..643bed0f
--- /dev/null
+++ b/RightView.py
@@ -0,0 +1,39 @@
+# 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 : Do level order traversal using a queue.
+# For each level, we add the last node's value to the result which gives the right side view of the tree.
+# For each level, if we add the first node's value to the resul, then it will give the left side view of the 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
+class Solution:
+    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
+        res = []
+
+        if root is None:
+            return res
+
+        q = deque()
+        q.append(root)
+
+        while q:
+            size = len(q)
+
+            for i in range(size):
+                curr = q.popleft()
+
+                if i == size - 1:
+                    res.append(curr.val)
+
+                if curr.left:
+                    q.append(curr.left)
+                if curr.right:
+                    q.append(curr.right)
+        return res
+        
\ No newline at end of file