Smallest String Starting From Leaf - Leetcode 988 - Python

[@jessicakoch2331]

even if I solve a problem, I immediately go to see your explanation, so many of leetcode’s editorials are indecipherable (at least for me)

[@tekfoonlim4745]

This is today's leetcode problem. Nice explanation man!

[@raviyadav2552]

can you make a series dedicated to trees and related concepts?

[@licokr]

Thanks for uploading the video! It motivates me. Consistency is the key! 🔑

[@adityamwagh]

Hey, I think it would be a good idea to make a webextension which enables us to see this video in the leetcode tab itself!

[@sathwikmadhula9527]

Neetcide, I request you to please solve the concurrency problems on leetcode. The leetcode numbers are 1114, 1115, 1116, 1117, 1188, 1195, 1226, 1242, 1279.

[@vishaalkumaranandan2894]

That usage of min() was brilliant!!

[@daikaji3833]

Today is the day. I'm tired of sitting here with this degree and being unable to use it. No more wasted potential. I am buying the 1 yr subscription right now, and I'm going all in to prepare for job hunting. Thank you for all that you do.

[@MiraKumar-rm5ke]

I have a question, for the input - [25, 1, 3, 1, 3, 0, 2] shouldn't the output be "bbz" and not "adz" (actual output). Your input is appreciated. thanks! @NeetCodeIO

[@fancypants6062]

I had to figure out the same thing as you. It is because the "lexicographically smaller" does not mean that the sum of the letters (treated as numbers) is smaller; it means alphabetic order, so anything starting with a is ahead of anything starting with b, no matter what the rest of the letters are.

[@sankhadip_roy]

Less code but same concept class Solution: def smallestFromLeaf(self, root: Optional[TreeNode]) -> str: self.ans="{" def dfs(node, word): if not node: return word = chr(97 + node.val)+word if not node.left and not node.right: if word

[@SC2Edu]

I used the same approach, it made more sense in my head.

[@sankhadip_roy]

@@SC2Edu Yes absolutely

[@MP-ny3ep]

Thank you so much. This was so beautifully explained.

[@mohammadmohtashim7985]

Man this is a good solution.

[@yang5843]

I wrote mine a little differently class Solution { String rc = ""; public String smallestFromLeaf(TreeNode root) { dfs(root,""); return rc; } void dfs(TreeNode root, String temp) { if ( root == null ) return; temp = (char) ('a'+root.val)+temp; if (root.left == null && root.right == null ) { if ( rc.equals("") || temp.compareTo(rc) < 0 ) rc = temp; return; } dfs(root.left,temp); dfs(root.right,temp); } }

[@og_23yg54]

dump way use E2AV way 77% faster

[@rafaelhung1812]

thank you for your consistency

[@krateskim4169]

Thank you so much

[@NursultanBegaliev]

Thank you! 👍👍

[@chrischika7026]

im pretty sure you dont need the if not root: return part because you never have node

[@PedanticAnswerSeeker]

table = [chr(ord('a')+i) for i in range(26)] def helper(root, nowStr): if root==None: return nowStr if root.left==None: return helper(root.right, table[root.val]+nowStr) if root.right==None: return helper(root.left, table[root.val]+nowStr) return min(helper(root.left, table[root.val]+nowStr), helper(root.right, table[root.val]+nowStr)) return helper(root, '')

[@billyrobins6515]

Nice one, good to know, that I am the first here

[@chien-yuyeh9386]

🎉🎉

[@dss963]

Now tell me which is smaller in this according to the problem, "jd" or "hud"

[@zweitekonto9654]

Greedy would work when we were reading from top down.

[@3ombieautopilot]

The question is a little bit too easy for the Medium level.

[@VinayKumar-jf3ue]

you're code used beat 80-90% of users

[@SC2Edu]

Why not just check which string is smaller when we are at a leaf node?

[@jayrathod9271]

On the first leaf node, we don't have information about other leaf nodes

[@I.II..III...IIIII.....]

Yeah, that's what I did. I initiated my minimumString as "", and then I just checked if minimumString == "" or currentString < minimumString then minimumString = currentString

[@yusareba]

may I ask what highlighting/writing software?

[@leeroymlg4692]

paint 3D

[@aazzrwadrf]

such a bad question tbh