All Nodes Distance K in Binary Tree | Approach-1 | Leetcode-863 | AMAZON | Explanation ➕ Live Coding

Hi everyone, this is the 29th video of our "Binary Tree" Playlist.
In this video we will try to solve a very famous Binary Tree Qn “All Nodes Distance K in Binary Tree”.
We will write very easy and clean code.
We will do live coding after explanation and see if we are able to pass all the test cases.
In this video, we will solve it using Approach-1 (Using Parent Pointers)
Problem Name : All Nodes Distance K in Binary Tree
Company Tags : Amazon, Citicorp, Flipkart, Goldman Sachs, Hike, Ola Cabs, Samsung, Walmart
My solutions on Github : github.com/MAZ...
Leetcode Link : leetcode.com/p...
My GitHub Repo for interview preparation : github.com/MAZ...
Subscribe to my channel : / @codestorywithmik
Instagram : / codestorywithmik
Facebook : / 100090524295846
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#coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge #leetcodequestions #leetcodechallenge #hindi #india #hindiexplanation #hindiexplained #easyexplaination #interview #interviewtips
#interviewpreparation #interview_ds_algo #hinglish

[@Devildriver4441]

crystal clear explanation.Thank you bhaiya

[@codestorywithMIK]

Thanks a lot Bishal ❤️

[@sachinmohanty4577]

I woke up in the morning attempted this ..i was so out of my mind ki mei queue soch ke stack ke hisab se pop kar raha hun dry run mei 😂😂 then finally samjha ki kitna bada blunder kar raha tha ... Thanks MIK ❤

[@codestorywithMIK]

🤓🤓 Thank you for watching ❤️❤️

[@wearevacationuncoverers]

You have the skill to make things look like a CAKE WALK ❤ I still am so glad that I found this channel

[@abc-ym4zs]

you have very unique skill to explain very good problems in a very understandable way only one request sir please continue these series of solving dialy problems sir finally a big thanks to you sir whenever i fell demotivated in DSA after seeing ur solutions again i am pumped up to solve questions sir

[@codestorywithMIK]

Thank you so much 🙏😇❤️

[@akshatdubey4421]

The way you explained this is so simple and intuitive. Thanks a lot

[@ntsequalifier341]

Really well explained.

[@AbhijeetMuneshwar]

So intuitive !!! 💡

[@pawankumaryadav8127]

Beautiful explanation 😊

[@kartikrameshchavan4710]

Thanks Sir 🙏

[@codestorywithMIK]

Thank you for watching ❤️🙏

[@viditvaish7317]

asusual no one can beat ypu in explaining the concept ans dsa question

[@krishnaharshith8120]

Hey I have been following ur videos to help me solve my daily leetcode problems, and the way u teach is impressive and it motivates me to do atleast one question everyday. Thanks fror the content you put out, hope you keep doing this and would love to get to interact with you , as a mentor or a guide coz i am pretty confused and unguided all the time. Once again thank you, if you are looking at this comment please do reply, would love to get to message you irl. Thank you

[@codestorywithMIK]

Thanks a lot 🙏😇❤️ You can connect on LinkedIn

[@ugcwithaddi]

Best explanation

[@xiaoshen194]

Thnk u for the soln. Main to pta nhi kya tatti approach laga rha tha level order traversal ke sath 🥲😞

[@floatingpoint7629]

thanks for the explanation. i was able to solve this one by self 😄

[@codestorywithMIK]

Wowieee. Awesome 😍😇🙏

[@killshot57]

Dp concept and qns series ko please continue kardo 🥺🥺

[@codestorywithMIK]

Yes yes.Coming today/tomorrow ❤️❤️🙏🙏

[@anuppatankar4294]

Great Explanation 👌🏻

[@codestorywithMIK]

Thank you so much Anup ❤️❤️

[@krishanapoorv7710]

I was waiting for your video since morning so that i can get more approaches ,this hash-map approach is everywhere but what if i have to solve this question without hash-map without using parent-child relationship and also not converting it into graph.so can you solve this with another approach. I have doubt in one code can you make me understand how that code is running class Solution { public: vector distanceK(TreeNode* root, TreeNode* target, int k) { vector res; dfs(root, target, k, res); return res; } private: int dfs(TreeNode* node, TreeNode* target, int k, vector& res) { if (!node) { return -1; } if (node == target) { subtreeAdd(node, 0, k, res); return 0; } int left = dfs(node->left, target, k, res); int right = dfs(node->right, target, k, res); if (left != -1) { if (left + 1 == k) { res.push_back(node->val); } subtreeAdd(node->right, left + 2, k, res); return left + 1; } if (right != -1) { if (right + 1 == k) { res.push_back(node->val); } subtreeAdd(node->left, right + 2, k, res); return right + 1; } return -1; } void subtreeAdd(TreeNode* node, int dist, int k, vector& res) { if (!node) { return; } if (dist == k) { res.push_back(node->val); } subtreeAdd(node->left, dist + 1, k, res); subtreeAdd(node->right, dist + 1, k, res); } };

[@user-og1ss2pv2l]

Awesome explanation 👍👍👍

[@codestorywithMIK]

Glad you liked it 👍🏻🙏

[@tutuimam3381]

Thanks a lot

[@dayashankarlakhotia4943]

Good explanation with dryrun

[@codestorywithMIK]

Thanks a lot 🙏😇

[@dikshantyadav1110]

Nice explanation, keep going

[@codestorywithMIK]

Thank you 🙂❤️

[@Tanishq181]

Thank you so much ❤

[@codestorywithMIK]

You're welcome 😊

[@GeneralistDev]

Sir my brute approach was to convert the tree into adj list form since each node->vak is unique . Is it not good for interview ?

[@codestorywithMIK]

That’s absolutely an acceptable approach. Also leetcode has put this in their official solution

[@Momentsofmagic28]

Made it a cake walk ❤❤

[@schrodingerskatt222]

It would be better to have a hashmap of TreeNodes rather than int, suppose you have repeated numbers at kth distance from target, then it would print it only once.

[@codeandtalk6]

❤

[@SaranshKasliwal]

@codestorywithMIK Can you upload Video on the other approach to solve this question

[@vineetkumar2899]

I think bhaiya interviewer will not be happy with this approach 😢.Please make video on second approach soon.

[@codestorywithMIK]

Indeed the interviewer will ask for better one. And that’s why I made a separate video on approach-1 Definitely I am gonna make Approach-2 video as well 😇❤️

[@floatingpoint7629]

whats wrong with this approach. this is the official approach chosen by leetcode itself

[@codestorywithMIK]

@floatingpoint7629 It’s not a bad approach. However, it can be solved without using map and parent pointer. So, in case interviewer asks to solve without using parent pointers.

[@floatingpoint7629]

@@codestorywithMIK got it. looking forward to that solution 😀

[@vineetkumar2899]

I'm waiting for that 😁

[@unodiscite]

inorder traversal ( left root node)

[@alphadrones23]

This was asked to me in Amazon interview

[@codestorywithMIK]

Wow 😯

[@souravjoshi2293]

Wow itna easy tha kya ye 😥

[@abinashdash7864]

Please make a video without using parent pointer

[@husler7424]

Can u please create Recursion playlist?

[@kamranwarsi12b22]

Can we store the parent mapping using level order traversal ??

[@codestorywithMIK]

Yes definitely

[@kamranwarsi12b22]

Ohkk , thnx

[@shivanshnegi6957]

bhaiya dp playlist kaafi time sai update nhi hui plz atleast 1 week mai 3 video daal diya kaaro dp ki 🥲🥲

[@codestorywithMIK]

Sure thing. Will upload this week ❤️❤️❤️

[@SohelKhan-vt5ql]

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void inorder(TreeNode *root , vector adj[]) { if(not root) return; if(root -> left) { adj[root -> left -> val].push_back(root -> val); adj[root -> val].push_back(root -> left -> val); } inorder(root -> left , adj); if(root -> right) { adj[root -> right -> val].push_back(root -> val); adj[root -> val].push_back(root -> right -> val); } inorder(root -> right , adj); } vector distanceK(TreeNode* root, TreeNode* target, int k) { vector adj[501]; inorder(root , adj); vector visited(501 , false); for(int i = 0 ; i < 3 ; i++) { cout

[@blackstargaming3678]

bhai please approach 2 jaldi le lao please request he

[@RajGupta-cu9hi]

Where is the approach 2

[@_singh516]

more intutive is Postorder

[@codestorywithMIK]

Thank you for sharing 😇