BFS-2 Done

Leetcode_199.java

@@ -0,0 +1,97 @@
+//way1
+/**
+ * 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;
+ *     }
+ * }
+ */
+ //BFS
+ //Travel level by level using queue.size
+ //Store last element
+ //TC: O(n);
+ //SC:O(n)
+class Solution {
+    public List<Integer> rightSideView(TreeNode root) {
+        List<Integer> ans=new ArrayList<>();
+        if(root==null){
+            return ans;
+        }
+
+        Queue<TreeNode> queue=new LinkedList<>();
+        queue.add(root);
+        while(!(queue.isEmpty())){
+            int size=queue.size();
+            for(int i=0;i<size;i++){
+                TreeNode curr=queue.poll();
+                if(i==size-1){
+                    ans.add(curr.val);
+                }
+
+                if(curr.left!=null){
+                    queue.add(curr.left);
+                }
+
+                if(curr.right!=null){
+                    queue.add(curr.right);
+                }
+            }
+        }
+
+        return ans;
+
+    }
+}
+
+//way2
+/**
+ * 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;
+ *     }
+ * }
+ */
+ //DFS
+ //Travel all nodes 
+ //At every level add the node to ans
+ //By end - we will be left with right view of ans
+ //TC: O(n)
+ //SC: O(h)
+class Solution {
+    public List<Integer> rightSideView(TreeNode root) {
+        List<Integer> ans=new ArrayList<>();
+        helper(root,0,ans);
+        return ans;
+    }
+
+    private void helper(TreeNode root,int level, List<Integer> ans){
+        if(root==null){
+            return;
+        }
+
+        if(ans.size()==level){
+            ans.add(root.val);
+        }else{
+            ans.set(level,root.val);
+        }
+
+        helper(root.left,level+1,ans);
+        helper(root.right,level+1,ans);
+    }
+}

leetcode_993.java

@@ -0,0 +1,132 @@
+//way1
+/**
+ * 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;
+ *     }
+ * }
+ */
+ //BFS
+ //Store root in queue
+ //Travel till the size of queue is met - level order traversal
+ //For every node check whether it's both left and right are equal to x and y --> if yes, siblings. Not cousion. Return false
+ //For every level check whether both x and y are found or not. As we have already checked and they are not siblings. So, they are cousins. Return true
+ //Once queue elements are processed but still x and y are not found in one level - return false
+ //TC: O(n)
+ //SC: O(n)
+class Solution {
+    public boolean isCousins(TreeNode root, int x, int y) {
+        Queue<TreeNode> queue=new LinkedList<>();
+        queue.add(root);
+
+        while(!(queue.isEmpty())){
+            int size=queue.size();
+            boolean xFound=false,yFound=false;
+
+            for(int i=0;i<size;i++){
+                TreeNode curr=queue.poll();
+
+                if(curr.left!=null && curr.right!=null){
+
+                    if((curr.left.val==x && curr.right.val==y) || (curr.left.val==y && curr.right.val==x)){
+                        return false;
+                    }
+                }
+
+                if(curr.val==x){
+                    xFound=true;
+                }
+
+                 if(curr.val==y){
+                    yFound=true;
+                }
+
+                if(curr.left!=null){
+                    queue.add(curr.left);
+                }
+
+                if(curr.right!=null){
+                    queue.add(curr.right);
+                }
+
+            }
+
+            if(xFound==true && yFound==true){
+                return true;
+            }
+
+        }
+
+        return false;
+
+    }
+}
+
+//way2
+/**
+ * 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;
+ *     }
+ * }
+ */
+ //DFS
+ //Travel every node
+ //At every node check it's children are equal to x and y. If yes, siblings and make result false
+ //Else, check if node.val is x or y. if yes, mark it's levels
+ //At recurssion, if levels are matching, return result
+ //if levels are not matching, return false;
+ //TC:O(n);
+ //SC:O(h)
+class Solution {
+    int xlevel=-1;
+    int ylevel=-1;
+    boolean result=true;
+    public boolean isCousins(TreeNode root, int x, int y) {
+        helper(root,x,y,0);
+
+        if(xlevel==ylevel) return result;
+        return false;
+
+    }
+    private void helper(TreeNode root, int x, int y, int level){
+        //
+        if(root==null){
+            return;
+        }
+
+        if(root.val==x){
+            xlevel=level;
+        }
+
+        if(root.val==y){
+            ylevel=level;
+        }
+
+        if(root.left!=null && root.right!=null){
+            if((root.left.val==x && root.right.val==y) || 
+            (root.left.val==y && root.right.val==x)){
+                result=false;
+            }
+        }
+
+        helper(root.left,x,y,level+1);
+        helper(root.right,x,y,level+1);
+    }
+}