leetcode

commit d416fa58f68935419f8c4b2a1c4e8a9ad529eeff
parent 6ec1cfdb822cf3e859a140a838fec9ec33101c35
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Wed, 10 May 2023 16:07:03 -0500

Completed binary tree zigzag level traversal problem using dart

Diffstat:
A
binary-tree-zigzag-level-order/zig-zag-level-order.dart
 | 
53
+++++++++++++++++++++++++++++++++++++++++++++++++++++

1 file changed, 53 insertions(+), 0 deletions(-)
diff --git a/binary-tree-zigzag-level-order/zig-zag-level-order.dart b/binary-tree-zigzag-level-order/zig-zag-level-order.dart
@@ -0,0 +1,53 @@
+//Traverse a binary tree and return a list that contains
+//one list for each layer. As you make your way down the tree
+//start going from left to right and then from right to left and so on.
+//The time complexity of this code is O(n) where n is the 
+//number of nodes in the tree.
+//Time: 264ms Beats: 63.64%
+//Memory: 143.1MB Beats: 54.55%
+
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *   int val;
+ *   TreeNode? left;
+ *   TreeNode? right;
+ *   TreeNode([this.val = 0, this.left, this.right]);
+ * }
+ */
+class Solution {
+  List<List<int>> zigzagLevelOrder(TreeNode? root) {
+      if(root == null){
+          return [];
+      }
+      List<TreeNode?> queue = [];
+      List<List<int>> return_list = [];
+      queue.add(root);
+      int row = 0;
+      while(queue.length > 0){
+          return_list.add([]);
+        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);
+            }
+            return_list[return_list.length - 1].add((queue[0]?.val ?? 0));
+            queue.removeAt(0);
+        }
+        row += 1;
+      }
+
+      for(int i = 1 ; i < return_list.length ; i += 2){
+          return_list[i] = return_list[i].reversed.toList();
+      }
+
+
+
+      return return_list;
+
+  }
+
+}