L30. Print all the Nodes at a distance of K in Binary Tree | C++ | Java

Entire DSA Course: takeuforward.org/strivers-a2z...
Check our Website:
Linkedin/Instagram/Telegram: linktr.ee/takeUforward
#treeSeries #striver #placements

[@takeUforward]

Please likeeee, shareeee and subscribeeeeeeee :) Also follow me at Insta: Striver_79

[@willturner3440]

Done brother this thing is obvious 😁

[@ayushjain386]

On 50 line ,i think wha pe parent_track[current]. second aana chaiye because we marked the parents like 5->3 5 ka parent 3 hai

[@PrinceKumar-el7ob]

@@ayushjain386 no it's correct 5->3 mtlb 5 ka parent 3 hai to parent_track[current] =3 hi aayega if current=5

[@JoshMartin47]

unordered_map in c++ does take O(1) time for look up. so the time complexity in worst case will come out to O(n)

[@soumyadeepdas1536]

Bhaiya for the mark parent function there is probably no need to carry the target node seperately ig?? coz we are just marking the parent nodes for the correspnding child nodes??

[@symbol767]

To perhaps make it more clear for those still a bit confused He basically turned a Binary Tree into an Undirected Graph, this method is incredible and extremely useful.

[@vivekmishra641]

100 bat ki 1 bat

[@someshpatel7660]

Why cant we use DFS and create a mod distance variable which set to 0 at node and going away increment by 1 and on return decrement by 1. When it again comes to node, on return of that node it will -1. Here we can use absolute distance. Thougths?

[@aakashpatel1250]

​@@someshpatel7660this won't work if the target is on the right subtree and the k distant node is on left subtree of the root. Correct me if I misunderstood

[@nopecharon]

What i learned: When you have to traverse back use a map to store the parent node

[@phatcat7924]

these are the comments i look for they straight away go into my revision notes.

[@prakhargupta5410]

@@phatcat7924 could you please share your notes with us🥺

[@tushar8579]

@@phatcat7924 Bro he is basically making a graph and doing bfs to reach all nodes at kth level.

[@pulkitjain5159]

yups

[@uRamPlus]

Self Notes: 🍋 Mark each node to its parent to traverse upwards 🍋 We will do a BFS traversal starting from the target node 🍋 As long as we have not seen our node previously, Traverse up, left, right until reached Kth distance 🍋 when reached Kth distance, break out of BFS loop and remaining node's values in our queue is our result

[@gautamarora6556]

Thank you

[@thinkingmad1685]

Your self notes help me as well 😄

[@shikharbansal2942]

Helpfull Thanks!

[@parthsalat]

Wherever I go...I search for your "self notes"

[@ashishdhal4614]

Teach me your ways senpai

[@himanshugupta7010]

We can implement it using recursion as well. As on every node , there will be 3 recursion .. i.e for left , for right and for parent .. code is given below :: void makeParent(TreeNode* root,unordered_map &parent){ queue q; q.push(root); while(!q.empty()){ int n= q.size(); for(int i=0;ileft) { parent[node->left]=node; q.push(node->left); } if(node->right){ parent[node->right]=node; q.push(node->right); } } } } class Solution { public: vector distanceK(TreeNode* root, TreeNode* target, int k) { unordered_map parent; makeParent(root,parent); unordered_map visited; vector ans; solve(target,parent,visited,k,ans); return ans; } void solve(TreeNode* target,unordered_map &parent,unordered_map &visited,int k,vector &ans){ if(k==0){ ans.push_back(target->val); } visited[target]=true; if(target->left && !visited[target->left]){ solve(target->left,parent,visited,k-1,ans); } if(target->right && !visited[target->right]){ solve(target->right,parent,visited,k-1,ans); } if(parent[target]!=NULL && !visited[parent[target]]){ solve(parent[target],parent,visited,k-1,ans); } }

[@shubhamuppal-1599]

after doing every question of this series i get to know that main motive is not to prepare for questions in interview/coding round but to identify pattern. Must say striver your content is top notch.

[@anujjain9273]

with this logic , i code code k distance node downwards and upwards, really impressed with the logic , dry run took time , i did it two times though , Thanks for making such content

[@shashankojha3452]

Thanks!

[@supratimbhattacharjee5324]

So basically we are traversing a tree as a graph and doing BFS from the given node

[@PrinceKumar-el7ob]

yeah forming a graph and doing BFS exactly.

[@aman5534]

Its kinda funny 😂

[@rushidesai2836]

Tree is a graph.

[@pulkitjain5159]

crux : converted tree into an undirected graph and applied a dfs / bfs .

[@pratyushnarain5220]

you can also do it in O(H) space by stroring root to target node path and then calling the k-down function on them.

[@nikhilmeena8585]

even we can do it without storing the root to that node path , by just checking whether if a nodes leftchild contains target , then we will search for possible answers in the right subtree of current node , and if found in rightNode then w will check possible answers in left subtree , if the node is itself target than we can just see all its childrens at distance k.

[@ishangujarathi10]

Superb Intuition and explanation, this problem falls in the range of Hard Problem, but your technique and approach makes it super easy to understand and also code!!

[@gandhijainamgunvantkumar6783]

What an amazing explanation. I was able to do the whole code by myself just after you did a dry run and told the logic. Thank you so much bhaiya for making trees easy for us. :)

[@_sf_editz1870]

here is the java code with target as integer and also target as a node thank you striver bhayya for making the concept clearer public class Solution { public static List distanceK(TreeNode root, int target, int k) { Map parent = new HashMap(); markParents(root, null, parent); Queue queue = new LinkedList(); Set visited = new HashSet(); TreeNode tgt = findNode(target , root); queue.offer(tgt); visited.add(tgt); int level = 0; while (!queue.isEmpty()) { if (level == k) break; int size = queue.size(); level++; for (int i = 0; i < size; i++) { TreeNode current = queue.poll(); if (current.left != null && !visited.contains(current.left)) { queue.offer(current.left); visited.add(current.left); } if (current.right != null && !visited.contains(current.right)) { queue.offer(current.right); visited.add(current.right); } TreeNode parentNode = parent.get(current); if (parentNode != null && !visited.contains(parentNode)) { queue.offer(parentNode); visited.add(parentNode); } } } List result = new ArrayList(); while (!queue.isEmpty()) { result.add(queue.poll().val); } Collections.sort(result); return result; } public static void markParents(TreeNode root, TreeNode par, Map parent) { if (root == null) return; parent.put(root, par); markParents(root.left, root, parent); markParents(root.right, root, parent); } static TreeNode findNode(int val , TreeNode root){ if(root==null) return null; if(root.val == val) return root; TreeNode left = findNode(val , root.left); TreeNode right = findNode(val , root.right); if(left==null) return right; if(right == null) return left; return null; } }

[@vaibhavgupta973]

starting mae toh ache samaj nhi aa rha tha . but jaise hi code walk through kra ... sab samaj aa gaya ache se. Thanks!!

[@BinduMahapatra-sq6vh]

Ek TVF and dusra TUF bas ye dono hee rocking hai abhi tou.

[@armaanhadiq3741]

Basically, here we are making the undirected graph from given tree and using BFS(level order traversal of graph) to find different vertices at distance k

[@pulkitjain5159]

yes

[@beamncrash9971]

yeah more like creating a adjaceny list and then doing BFS from target node

[@harshmittal3128]

Thanks for such a great explanation striver... This was a very good question , learnt multiple new approaches from this one ..

[@cinime]

Understood! Such a fantastic explanation as always, thank you very much!!

[@gouravkumar7459]

The easiest explanation for this problem so far on youtube.

[@Anonymous-uj3jx]

Why everything becomes soo easy when striver teaches it ? Mann you are magical 💖

[@AppaniMadhavi]

ya

[@solankiketul5640]

I have one more solution with time complexity O(2n) and space complexity O(n). 1.) Take a map which stores the pair (Node value, direction from root, left or right), eg, (4, left) 2.) in the process, store the level of the target and the direction, level=1, direction= left 3.) if the target is on left, take all the values from level+k with direction left. In our eg, from map[3], we will get 7 and 4 4.) now for ancestor, take map[k-level], so we will have map[1] and as the target value is on left, we will take the nodes with direction right from map[1]. In our example it is 1.

[@stith_pragya]

Thank You So Much for this wonderful video...........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[@willturner3440]

For me this is the most awaited video.. Love you striver 😍

[@charlesbabbage6786]

What a mind blowing solution!!

[@paritoshdadhich2954]

Thank you for the best possible solution. Hats off to your efforts

[@NeerajSharma-mz4es]

I leanned a new approach thanks to u sir

[@deepakjain4481]

i think for this method we should have used 3 pointers in a binary tree left right and parent while constructing tree and then simply traverse the tree and finding the target of the tree and then using a map i which two variable are there int for distance and node for distinct element

[@tusharnain6652]

There is no need of passing target to makr parent function.

[@abhineetkapil]

Nice and easy explaination. !

[@palakmantry]

Thank you for the explanation, really helpful

[@teja1384]

if we have to traverse a tree upward then we have to make parent map which store the parent of every node , Nice explanation.

[@emmatime2016]

Soooooo clear !!!! Thank you!

[@vanshajtiwari1282]

Love you bhai love you, You are amazing, amazing explaination.

[@silicon9794]

Good explanation bro. Understood properly. Thankyou :)

[@iWontFakeIt]

here, the problem in leetcode has constraints given that 0

[@bhashkarbelwal4116]

you are amazing tutor #takeuForward bro

[@rajatrajgautam3224]

I basically learned if we want to go backward in a tree we need to use a map to store the parent node.........Incredible !

[@user-es4vn2uw5l]

You are magic striver and you create magic

[@ayushidalal5488]

Amazing explanation! Thankyou so much :)

[@sikandarchaudharyzx9014]

Sometimes you make the easy questions very complex.

[@samyakjain7422]

just love ur videos man...great explanation...:)

[@gubbalamalleswari5892]

Super explanation

[@rishabhkumar8115]

Very well explained bhaiya! understood🔥

[@ishachauhan6477]

very well explained thankyou sir

[@riteshkhadse4517]

to mark visited nodes we can use set instead of map.

[@aakriti1]

Understood, Great explanation! 🤩

[@amisha2545]

🔥couldn’t have been better!

[@tanishq2766]

I solved it differently(return all the downward nodes at a distance k from some particular node and some simple manipulations), but i think this approach is easier to come up with if someone have studied standard graph problems

[@satyamsrivastava9034]

I encountered this problem in one of the interview I told him the approach which you have explained then he told me not to use the map to store the parents .. and then I shattered as I don't have that approach in mind. :(

[@mukulsaluja6109]

U can store root to target node path

[@saikrishnalanka133]

In the first step you can find the nodes which are at k distance below given node using recursive traversal. Then back track to each ancestor and store distance of ancestor in variable ( say d). Back track until k-d!=0. and at each ancestor call recursive again to find node at distance k-d.

[@chirayubaliyan5181]

what about pair, you can use that also!Maybe!

[@aryashjain7893]

Ill do this again , accha problem hai revision lagega thenks striver

[@nagavedareddy5891]

Perfect and most efficient explanation.. But small optimisation would be using hash set instead of Hashmap for visited.

[@chandrachurmukherjeejucse5816]

I solved it myself by another approach. Please let me know If It is a good approach or not. 1. Storing the path from root the the node in a deque using a dfs. 2. keep a count for how many elements are popped from deque 2. pop items from the front of deque and find the nodes at a dist (n - no of items popped). 3. Now to make sure that the latter popped node doesnot searches in the direction of the node popped previously we use a unordered set of popped out elements and after finding nodes at a dist for a node we put the node in the unordered set. Code: class Solution { private: bool dfs(TreeNode* root, TreeNode* target, deque &dq) { if(root == NULL) return false; if(root == target) { dq.push_back(target); return true; } bool isFound = false; isFound = dfs(root -> left, target, dq); isFound = isFound || dfs(root -> right, target, dq); if(isFound) { dq.push_back(root); return true; } return false; } void getAllNodes(TreeNode* curr, unordered_set &s, int dist, vector &res) { if(curr == NULL) return; if(s.find(curr) != s.end()) return; if(dist == 0) { res.push_back(curr -> val); return; } getAllNodes(curr -> left, s, dist - 1, res); getAllNodes(curr -> right, s, dist - 1, res); } public: vector distanceK(TreeNode* root, TreeNode* target, int k) { deque dq; unordered_set s; dfs(root, target, dq); vector res; int dist = k; while(!dq.empty()) { TreeNode* curr = dq.front(); dq.pop_front(); getAllNodes(curr, s, dist--, res); s.insert(curr); if(dist < 0) break; } return res; } };

[@051-avnee4]

Awesome explanation 💫💫!!! Understood .....

[@Abhishek-do8mp]

just one word - amazing

[@theghost9362]

ommmg , thanks , I was doing something completely different , I was trying to compute a list that represent that BT , and then compute the the parent and children K times until i get the result , but you're algo is a lot faster

[@4747surya]

Basically convert tree into a undirected graph start from target and do k iteration of BFS ?

[@user-tk2vg5jt3l]

Thank you Bhaiya

[@sukhpreetsingh5200]

Amazing explanation😇

[@nihalnandannayak8869]

What is the need of passing target node as argument of the function markparen()

[@gigglezone3432]

I solved it by converting tree to graph, and take adjacency list and bfs to traverse to all nodes at a distance of k? Is it a good approach for interviews?

[@rijumondal6876]

Do I need to go with a brute-better-optimal solution for this approach in the interview if asked??

[@Yash-uk8ib]

I initiallly thought of actually converting this tree to an undirected graph and just finding K distant nodes but your observation is just best!! Can u tell me why u took visited array??

[@takeUforward]

So that I dont go back to them..

[@Yash-uk8ib]

@@takeUforward sir u can visit back only parent nodes right? And that can be maintained using a variable?

[@deepaktiwari7059]

@@Yash-uk8ib We can also revisit the target node from parent node(for eg 3 to 5)

[@infinityzero2321]

I am thinking of anotger approach which is treating this like a directed graph. So instead of marking that whuch is the nodes parent we can just make an adj list and mark them like an undirected graph. Then we can directly do the bfs

[@mansisethi8127]

Dope question

[@pratikdas1780]

this is outright amazing. wow! i'm amazed

[@pratikdas1780]

bro in the past, it's a simple bfs graph traversal technique after you link all the parent nodes. still, it's amazing. just not as hard.

[@subhankarpal2260]

Understood... thanks a lot.

[@ajitheshgupta3017]

If it was print all from lead node. What changes has to be done?

[@suryakiran2970]

Great Explanation

[@sanchitdeepsingh9663]

thanks sir

[@radharamamohanakunnattur3035]

Understood!! awesome take

[@harshjhunjhunuwala]

Actually there's no need of "target" parameter in markParent function as it isn't used anywhere!

[@adityaagarwal2324]

Best explanation 🔥🔥

[@AlokKumar-ld9qr]

amazing explanation. 😃😃😃

[@harshitjaiswal9439]

understood.

[@jothsna]

Nice explanation☺

[@lavanyaprakashjampana933]

we love your content and we love you.....🖤

[@vedantsharma5876]

Should this(or a similar) question be asked in an interview? Because it took me almost 1 hour to just code!

[@amanbhadani8840]

Yes definitely,You have to work on your speed.

[@utkarshsharma6650]

understooood.thanks :)

[@bhaveshkumar6842]

Thank you bro!!!!

[@santanu29]

Great explanation

[@himanshidafouty347]

Understood

[@chiragbansod8252]

understood

[@pritamsaha7129]

What happens when we get two nodes of same value?? Will unordered map work at that time??

[@DevanshuAugusty]

why are you passing the target node int mark parent function?

[@YashKakrecha]

Wow nice explanation

[@littleblackhat2142]

How do we solve this without using parent pointers or without any extra space? Is it even possible? I was asked this question in Amazon interview and the interviewer was persistent on solving without space and ultimately it led to me bombing the interview :(

[@manojg4451]

Love the concept of converting tree to graph, Please do a playlist for bits and sorting

[@SUMITKAUSHIK1000]

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List distanceK(TreeNode root, TreeNode target, int k) { Map map = new HashMap(); Set visited = new HashSet(); List list = new ArrayList(); updateParent(null, root, map); traverse(k, target, map, list, visited); return list; } public void updateParent(TreeNode parent, TreeNode current, Map childToParent){ if(current == null) return; childToParent.put(current, parent); updateParent(current, current.left, childToParent); updateParent(current, current.right, childToParent); } public void traverse(int distance, TreeNode node, Map childToParent, List list, Set visited){ if(node == null) return; if(visited.contains(node)) return; visited.add(node); if(distance == 0){ list.add(node.val); return; } traverse(distance-1, node.left, childToParent, list, visited); traverse(distance-1, node.right, childToParent, list, visited); traverse(distance-1, childToParent.get(node), childToParent, list, visited); } }

[@parthsalat]

Please do a smooth transition from black iPad to white Code screen...I've become blind because of the quick transition 😥

[@rishabsharma5307]

please make videos on finding time complexity of finding complex questions

[@takeUforward]

Comes with practice, don't think on that too much :)

[@shivampratapsingh2188]

I have one doubt -this above apprach will get fail when tree will not be a binary tree right??

[@alesblaze4745]

thanks mate!

[@sakshisinghal1669]

Where are we incrementing cur_level after done with the level? Can anyone help?

[@prekshaswami3142]

why have we taken target as one of the argument in markParents function?

[@pritishpattnaik4674]

Just Excellent

[@randeepsaini5491]

very helpful

[@pranavsharma7479]

simple bfs traversal on the graph

[@agrawalmitesh4395]

GEM

[@NoName-hd7ok]

Store parents and do normal bfs using visited array