leetcode

commit 7760d9501adef9876c9ddaf7ca025ab7f47f4c55
parent d416fa58f68935419f8c4b2a1c4e8a9ad529eeff
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Wed, 10 May 2023 17:50:19 -0500

Completed binary tree right side view problem using dart and BFS

Diffstat:
A
binary-tree-right-side-view/binary-tree-right-side-view.dart
 | 
42
++++++++++++++++++++++++++++++++++++++++++

1 file changed, 42 insertions(+), 0 deletions(-)
diff --git a/binary-tree-right-side-view/binary-tree-right-side-view.dart b/binary-tree-right-side-view/binary-tree-right-side-view.dart
@@ -0,0 +1,42 @@
+//Given the root of a binary tree return all of the values
+//contained within the nodes if someone were looking at the tree from the right side.
+//Ex:
+//      1 <-
+//     2 3 <-
+//    3 4 <-
+//   6 <-
+//    4 <-
+//The time complexity of this code is O(n) where n is the
+//number of nodes in the list. My solution uses BFS and after
+//each layer is traversed it returns the rightmost node's value.
+//Time: 293ms Beats: 50%
+//Memory: 145MB Beats: 100%
+
+class Solution {
+  List<int> rightSideView(TreeNode? root) {
+    if(root == null){
+      return [];
+    }
+      List<TreeNode?> queue = [];
+      queue.add(root);
+      List<int> return_list = [];
+      return_list.add(root?.val ?? 0);
+      while(queue.length != 0){
+        int init_length = queue.length;
+        for(int i = 0 ; i < init_length ; ++i){
+          
+          if(queue[0]?.left != null){
+            queue.add(queue[0]?.left);
+          }
+          if(queue[0]?.right != null){
+            queue.add(queue[0]?.right);
+          }
+          queue.removeAt(0);
+        }
+        if(queue.length != 0){
+          return_list.add((queue[queue.length - 1]?.val ?? 0));
+        }
+      }
+      return return_list;
+  }
+}