LeetCode Invert Binary Tree Solution Explained - Java

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[usptact]

Eight years as a software engineer, I finally know how to invert a binary tree (and what it even means).

[zacisrael]

Wow, I can't believe I thought this was way more complex than it actually was. Thank you.

[Ohmriginal]

“Do you know what the scariest thing in the world is?”

[pooblock]

The explanation part for the code wasn’t the best haha but I figured it out. Not a bad video but try teaching things like we are stupid.

[uni]

This is deceptively simple. I can't believe I've been avoiding this particular problem for so long. I'm off to find similar problems so I can practise!

[兴宣大院君-hs]

It was mentioned in the leetcode solution BFS approach would save stack overhead to be more scalable

[ryanmichaud]

Thanks for the explanation! I’m a little confused why we don’t need to save the original right child value. I was thinking when we say root.right=left that this would make it so that both children hold the value of the left child. So on the following line when we say root.left=right we’re essentially setting left to left. Thanks again!

[AviPars]

Same as reflecting it ? Or mirroring ?

[alexander]

But one must first understand recursion in order to understand recursion…

[yu]

Just wondering, how does it go back up to 2 if it's DFS? It seems like it ends at 1 so I'm not sure how it travels back up. Please help😢

[OverLordOfDardWorld]

Holy shit, thanks Nick! Now I know how to invert a binary tree :D

[edino]

I think following could work as well, if tree is balanced then use array to store nodes, then reverse 2^k elements, so 4 -> 4, 27 -> 72 and 1369 -> 9631 and so on.
4 - 27 - 1369
4 - 72 - 9631

[catharsis]

The explanation made sense. Only 150 IQ could comr up with problem without looking at Algorithm ahead of time

[ThabiraGamingShorts]

Thank you Nick for such an explanation

[lucassmith]

Your code is fucking as intuitive as your logic! Love it! I Love you Nick!

[oscaropdyou]

This is slightly simpler. Top down invert.
private static void invert(TreeNode current) {
if (current == null) {
return;
}
TreeNode left = current.left;
current.left = current.right;
current.right = left;
invert(current.left);
invert(current.right);
}

[ipodtouch]

This is my bunk ass solution(first thing that I came up with)

public TreeNode invertTree(TreeNode root) {
if(root == null || (root.left == null && root.right == null))
return root;

// swap
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;

// move to next nodes
invertTree(root.left);
invertTree(root.right);

return root;

}

[vijaykumarreddyalavala]

Would you like to talk about the Inception thing happened at the beginning of the video ?

[Dmetal]

yeah ya know itll just ya know go back up ya know set left ya know to right and ya know ill explain that later ya know and itll hit this statement ya know ok thanks guys

[IbytheGOAT]

I hate the syntax of how to deal with these trees, I'm so used to things like arrays and strings and iterating through them using for loops and while loops that this feels completely brand new