Smallest String Starting From Leaf - Leetcode 988 - Python

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)

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

Thank you so much. This was so beautifully explained.

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

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

thank you for your consistency

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.

Thank you! 👍👍

That usage of min() was brilliant!!

Thank you so much

Man this is a good solution.

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!

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

🎉🎉

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

may I ask what highlighting/writing software?

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.

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

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

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); } }