From 5668b281a22631cc483e6a1e1a90abbdbc1f4104 Mon Sep 17 00:00:00 2001
From: sainathek <erukullasainath@gmail.com>
Date: Wed, 11 Mar 2026 15:05:49 -0400
Subject: [PATCH] Done BFS-2

---
 problem1.java | 68 +++++++++++++++++++++++++++++++++++++++++++
 problem2.java | 80 +++++++++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 148 insertions(+)
 create mode 100644 problem1.java
 create mode 100644 problem2.java

diff --git a/problem1.java b/problem1.java
new file mode 100644
index 00000000..0d5aa801
--- /dev/null
+++ b/problem1.java
@@ -0,0 +1,68 @@
+   /**
+    Time Complexity : O(N)
+    Explanation:
+    We perform a level-order traversal (BFS) and visit each node once.
+
+    Space Complexity : O(N)
+    Explanation:
+    The queue may contain up to one level of the tree at a time.
+    In the worst case (complete tree) it can store about N/2 nodes.
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially unsure how to capture the rightmost node of each level.
+    Solved it by performing BFS level by level and storing the last
+    node encountered in each level traversal.
+    */
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+
+class Solution {
+
+ 
+
+    public List<Integer> rightSideView(TreeNode root) {
+
+        Queue<TreeNode> q = new LinkedList<>();
+        List<Integer> li = new ArrayList<>();
+
+        if (root == null) return li;
+
+        q.add(root);
+
+        while (!q.isEmpty()) {
+
+            int size = q.size();
+            int last = 0;
+
+            for (int i = 0; i < size; i++) {
+
+                TreeNode t = q.poll();
+                last = t.val;
+
+                if (t.left != null) q.add(t.left);
+                if (t.right != null) q.add(t.right);
+            }
+
+            // The last node processed at this level is the rightmost
+            li.add(last);
+        }
+
+        return li;
+    }
+}
\ No newline at end of file
diff --git a/problem2.java b/problem2.java
new file mode 100644
index 00000000..a46d67d0
--- /dev/null
+++ b/problem2.java
@@ -0,0 +1,80 @@
+   /**
+    Time Complexity : O(N)
+    Explanation:
+    We traverse the tree once using DFS to locate both nodes
+    and record their parent and level.
+
+    Space Complexity : O(H)
+    Explanation:
+    Recursion stack can go up to the height of the tree (H).
+    In worst case (skewed tree) it can be O(N).
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially confused about the exact condition for cousins.
+    Later clarified that two nodes are cousins if:
+        1) They are at the same level
+        2) They have different parents
+    So while traversing the tree, I stored parent and level
+    for both nodes and compared them at the end.
+    */
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+
+class Solution {
+
+    int xparent = 0;
+    int yparent = 0;
+    int xlevel = 0;
+    int ylevel = 0;
+
+    public boolean isCousins(TreeNode root, int x, int y) {
+
+        dfs(root, 0, x, y);
+
+        if (xlevel == ylevel) {
+            return xparent != yparent;
+        }
+
+        return false;
+    }
+
+    private void dfs(TreeNode root, int level, int x, int y) {
+
+        if (root == null) return;
+
+        // Check if x is child of current node
+        if ((root.left != null && root.left.val == x) ||
+            (root.right != null && root.right.val == x)) {
+
+            xparent = root.val;
+            xlevel = level + 1;
+        }
+
+        // Check if y is child of current node
+        if ((root.left != null && root.left.val == y) ||
+            (root.right != null && root.right.val == y)) {
+
+            yparent = root.val;
+            ylevel = level + 1;
+        }
+
+        dfs(root.left, level + 1, x, y);
+        dfs(root.right, level + 1, x, y);
+    }
+}
\ No newline at end of file