Count of good nodes | Leetcode

[@johntakerChiku]

just at 2:29 i understood the concept. You have some magic bro. I didnt see the whole video, it just clicked at that particaluar time..thanks

[@CostaKazistov]

Please drop 60fps to 30 for high res uploads (720p and 1080p).
There is no need for super high refresh rate for tutorials.
It lags badly when jumping between parts of the video.

[@Pegasus02Kr]

Thanks for the great lecture as always!

[@techdose4u]

Thanks 😊

[@charanyasamudrala6724]

in order to maintain max at each node we need some datastructure with out searching whole path again and again.
i think max stack will give optimal time.
correct me if I'm wrong.😊

[@hritikrastogi3335]

anybody who want still less cody approach ... btw it is same as explained in video
void call(TreeNode* root,int &count,int nx)
{
if(root->val>=nx)
{
count++;
nx=root->val;
}
if(root->left) call(root->left,count,nx);
if(root->right) call(root->right,count,nx);
}

int goodNodes(TreeNode* root) {
int count =0;
call(root,count,-1e4);
return count;
}