Count Complete Tree Nodes | LeetCode 222 | C++, Java, Python

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June LeetCoding Challenge | Problem 23 | Count Complete Tree Nodes | 23 June,
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[@rashmikiranpandit8962]

Time complexity?
Interested to know which software do u use to explain and write on

[@manpreetkhokhar5318]

Loved the optimization part.I thought it is just a simple traversal problem.

[@rmadhavmca1]

Please talk about time complexity after the code complete.

[@devalmodi5350]

Very nicely explained!!

[@KnowledgeCenter]

Thank you 🙂

[@kedaramnaik5622]

Thanks for the explanation. It nice and easy to understand.
Can you tell me the software you use to create these videos?

[@KnowledgeCenter]

powerpoint

[@praveenj3112]

The simple recursive solution : as follows
If(root==null) return 0;
Return 1+countNodes (root.left)+countNodes (root.right);

[@KnowledgeCenter]

As I said in the video, any traversal will work. But, we should optimize the solution by utilizing Completeness property whenever possible.
For complete Binary Tree with all levels filled, it will take O(logn) time, bu for a normal Binary Tree, it will be O(n).
So, check if a given subtree is complete and filled, return 2^h - 1, else keep traversing.

[@praveenj3112]

@@KnowledgeCenter it's really good approach . thanks

[@pahehepaa4182]

pbs.twimg.com/media/DgG54TGUcAAKAUE.jpg Sorry I had too 😂

[@nativelearning]

Binary search is more optional one

[@subham-raj]

*A faster approach*
_As you don't require two traverse twice in every recursive call_
public class Solution {
public int CountNodes(TreeNode root) {
if(root == null)
return 0;

TreeNode leftNode = root, rightNode = root;
int height = 0;

while(rightNode != null) {
height++;
rightNode = rightNode.right;
leftNode = leftNode.left;
}

if(leftNode == null)
return (1

[@ijaz2020]

consider tree with no left node it has all node on right. then this code will throw NPE, right?