BFS-2

Cousins.py

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+# 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

RightView.py

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+# 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
+