level order traversal brute force

Author: kumar91gopi

leetcode/binary_tree_level_order_traversal.rb

@@ -4,6 +4,7 @@
 =end
 
 #Solution (Using BFS/Queue)
+# Time Complexity: O(n) Space: O(n)
 
 # Definition for a binary tree node.
 # class TreeNode
@@ -35,3 +36,36 @@ def level_order(root)
     end
     val_list
 end
+
+# Solution (Brute Force)
+# Time Complexity: O(n^2) Space: O(n)
+
+def level_order(root)
+    h = height(root)
+    trav_list = []
+    for i in 1..h
+        nodes = []
+        level_order_nodes(root,i,nodes)
+        trav_list.append(nodes)
+    end
+    trav_list
+end
+
+def level_order_nodes(root,level,nodes)
+    return nodes if root.nil?
+    if level == 1
+        nodes.append(root.val)
+    elsif level>1
+        level_order_nodes(root.left,level-1,nodes)
+        level_order_nodes(root.right,level-1,nodes)
+    end
+end
+
+
+
+def height(root)
+    return 0 if root.nil?
+    left_h = height(root.left)
+    right_h = height(root.right)
+    return 1+[left_h, right_h].max
+end