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  1. Practice
  2. Data Structures
  3. Trees
  4. Tree : Top View
  5. Discussions

Tree : Top View

  • aravindprasad + 2 comments

    Dont agree with this problem being tagged as "Easy".

    • maximshen + 5 comments

      I actually disagree with your disagree.

      "Top view" is simply defined as, "Starting from top node, every node added to the right side of top node is every right side node's right node if any, every node added to the left side of top node is every left side node's left node, if any." I just used a linkedlist to append left and right nodes on both sides from center to two ends.

      This problem has been completely misleaded by @Faland, unfortunately. And I don't think I am smarter than many folks here, so I kinda curious why this one caused so many confusion.

      • rajatbarman + 0 comments

        It caused confusion because GFG defines top view in a different manner than this question. http://www.geeksforgeeks.org/print-nodes-top-view-binary-tree/

      • Wafflenaut + 0 comments

        I think solving it is easy, but finding an elegant solution seems difficult. Not even sure how you'd do it without adding additional functions or creating something fairly icky.

      • hackmeifyoucan + 0 comments

        I just read your comment and solved it in 60 sec. Thanks very much :)

      • BansheeGhoul + 5 comments

        This question is just about printing the outermost left or right nodes .. java code (its a version of what parmarabhishek_1 told https://www.hackerrank.com/challenges/tree-top-view/forum/comments/331313 )

        java code

        void travLeft(Node root){
    if(root.left!= null)
          { travLeft(root.left);}
       System.out.printf("%d ",root.data);
               }
    void travRight(Node root){
     System.out.printf("%d ",root.data);
     if(root.right!= null){
         travRight(root.right);}
    }

   void topView(Node root) {
       if(root.left != null){
                    travLeft(root.left);
                            }
          System.out.printf("%d ",root.data);
       if(root.right != null){
                    travRight(root.right);
                            }      
    }
        • h415074476 + 1 comment

          you are right.

          • BarryB + 0 comments

            Actually the problem being tested with the test cases is easy. However if you try to solve the problem as defined in GFC, I'd agree that it's not that easy. I actually spent some time trying to solve what GFC describes, but I couldn't get all the test cases to work. When I debugged some of the test cases, I realized that the test cases were all for the simplified case. I then modified my code to support that, and all test cases passed.

            After I got all the test cases to work, I actually looked at the editorial, and found the following.

            Note: This solution is overly simplified and you will need something more complicated for certain types of imbalanced trees.

            Basically the actual problem statement was not well defined. Either the test cases should support all types of trees, and the problem could potentially be made a medium problem. Or the note should have been put in the problem statement, instead of the editorial.

        • mvbelgaonkar + 0 comments
          [deleted]
        • adityaverma82191 + 0 comments

          this will only pass the 1st test case. i thougth the same logic and i implemented it in c++.

        • kyxap + 0 comments

          Will not pass the 2nd TCs :)

        • ioluwayo + 0 comments

          This only passes the sample and first test case

      • jschopick + 0 comments

        I did this using helper functions and recursion but I printed the left values before the recursive call so it printed in reverse order. I thought I was just fundamentally wrong but I read this comment and then looked at my code in more detail and I just printed after the recursive call and it worked. Thank you!

    • tvanloon + 1 comment

      took me two minutes to solve

      • edemohankrishna1 + 1 comment

        can you please send me the code Mr tvanloon.I am basic coder please....:)

        • iam_ghanshyam + 0 comments

          Its better(learning-wise) to ask for a problem-solving approach instead of solutions.