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I came up with a similar solution in Java :) So far, I like it the most.
void top_view(Node root) { top_view(root, 0); } void top_view(Node root, int side) { if (root != null) { if (side <= 0) { top_view(root.left, -1); } System.out.print(root.data + " "); if (side >= 0) { top_view(root.right, 1); } } }
Your are very good!! Perfect Solution
Wow ... that is such a genius, elegant, fast, space efficient algorithm.
And I was failing in trying to adapt a top-view algorithm with a queue helper.
Woah, the simplest and best solution to this problem! I didnt think of this that way...
Very clever. :) Thank you.
What's the logic you used in order to develop it? Thank you!
I independently came up with a similar solution for Python:
def topView(root,side=0): if root is None: return 0 if side == 0 or side == -1: topView(root.left,side=-1) print root.data, if side == 0 or side == +1: topView(root.right,side=+1) return
this solution is amazing and graceful
Can someone please explain this code? Thank you
passes only one testcase
Tree : Top View
You are viewing a single comment's thread. Return to all comments →
I came up with a similar solution in Java :) So far, I like it the most.
Your are very good!! Perfect Solution
Wow ... that is such a genius, elegant, fast, space efficient algorithm.
And I was failing in trying to adapt a top-view algorithm with a queue helper.
Woah, the simplest and best solution to this problem! I didnt think of this that way...
Very clever. :) Thank you.
What's the logic you used in order to develop it? Thank you!
I independently came up with a similar solution for Python:
this solution is amazing and graceful
Can someone please explain this code? Thank you
passes only one testcase