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

[@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

[@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

[@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); } }

[@_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; } }

[@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.

[@shashankojha3452]

Thanks!

[@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.

[@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. :)

[@Anonymous-uj3jx]

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

[@AppaniMadhavi]

ya

[@vaibhavgupta973]

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

[@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

[@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!!

[@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.

[@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 !

[@gouravkumar7459]

The easiest explanation for this problem so far on youtube.

[@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

[@SibiRanganathL]

I just coded the solution with the explanation . THANK YOU STRIVER BHAI

[@5076_GidijalaDineshSurya]

Guys, this question was asked in Amazon interview. The twist is that we should not use a map to store the parents. Try solving it without using the map! Loved the explanation Striver!!

[@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.

[@sharmanihal99]

Python Code to Find all Nodes a K Distance in Binary Tree: #Thanks for the Great Explanation class Solution: def distanceK(self, root: TreeNode, target: TreeNode, k: int) -> List[int]: # Function to perform breadth-first search to find parent nodes def bfs_to_find_parents(root): parent_map = {} # Maps a node's value to its parent node if not root: return parent_map queue = deque() queue.append(root) while queue: node = queue.popleft() if node.left: queue.append(node.left) parent_map[node.left.val] = node # Store parent for left child if node.right: queue.append(node.right) parent_map[node.right.val] = node # Store parent for right child return parent_map # Function to find nodes at distance k from the target node def find_nodes_with_distance_k(target, parent_map, k): queue = deque() queue.append(target) visited = set() visited.add(target.val) distance = 0 while distance != k: size = len(queue) while size: node = queue.popleft() # Check if parent exists and it hasn't been visited before if parent_map[node.val] and parent_map[node.val].val not in visited: queue.append(parent_map[node.val]) # Add parent to queue visited.add(parent_map[node.val].val) # Mark parent as visited # Add left child to queue if it exists and hasn't been visited if node.left and node.left.val not in visited: queue.append(node.left)# Add left child to queue visited.add(node.left.val)# Mark left child as visited # Add right child to queue if it exists and hasn't been visited if node.right and node.right.val not in visited: queue.append(node.right)# Add right child to queue visited.add(node.right.val)# Mark right child as visited size -= 1 distance += 1 return queue # Build parent map parent_map = bfs_to_find_parents(root) parent_map[root.val] = None # Set root's parent as None # Find nodes at distance k from target nodes_at_distance_k = find_nodes_with_distance_k(target, parent_map, k) # Extract values of nodes at distance k result = [node.val for node in nodes_at_distance_k] return result

[@sikandarchaudharyzx9014]

Sometimes you make the easy questions very complex.

[@willturner3440]

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

[@BinduMahapatra-sq6vh]

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

[@divyanksharma236]

accha

[@rohan8758]

Key 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.

[@NeerajSharma-mz4es]

I leanned a new approach thanks to u sir

[@charlesbabbage6786]

What a mind blowing solution!!

[@haripriya8034]

we also do the problem like keeping a hashtable for distance and find all the distances from the root node for the left side... and for the right side as seperate.. and then we can find the nodes with that distance.

[@ritikshandilya7075]

Thankyou so much for great explanation Striver

[@stith_pragya]

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

[@riteshkhadse4517]

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

[@vaibhav56]

Solution without using any extra space for storing the parent: class Solution { public: vector ans; void down(TreeNode* target,int k){ if(!target)return; if(k==0){ ans.push_back(target->val); return; } down(target->left,k-1); down(target->right,k-1); } bool up(TreeNode* root,TreeNode* target,int &k){ if(!root)return false; if(root==target)return true; if(up(root->left,target,k)){ k--; if(k==0)ans.push_back(root->val); down(root->right,k-1); return true; } if(up(root->right,target,k)){ k--; if(k==0)ans.push_back(root->val); down(root->left,k-1); return true; } return false; } vector distanceK(TreeNode* root, TreeNode* target, int k) { down(target,k); up(root,target,k); return ans; } };

[@ayushgautam8000]

GOD Level Explanation 🫡🫡

[@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

[@aryashjain7893]

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

[@Yag116]

Solution :where we are given value of target Node void markparent(Node* root , unordered_map &keep_parent,int t,Node* &tn ){ if(!root) return; queue q; q.push(root); while(!q.empty()){ int n = q.size(); for(int i=0;idata==t) tn=root; //only for getting target node form given key if(root->left){ keep_parent[root->left]=root; // root k left ka parent root mark kr diya q.push(root->left); } if(root->right){ keep_parent[root->right]=root; // root k right ka parent root mark kr diya q.push(root->right); } } } } vector KDistanceNodes(Node* root , int target , int k){ unordered_map Keep_parent ; Node* targetN =NULL; markparent(root,Keep_parent,target,targetN); // sare parent aa gye unordered_map visted; queue q; q.push(targetN); visted[targetN] = true; int curr_dist=0; while(!q.empty()){ int size = q.size(); if(curr_dist++ == k) break; for(int i=0;ileft && !visted[curr->left]) { q.push(curr->left); visted[curr->left]=true; } if(curr->right && !visted[curr->right]) { q.push(curr->right); visted[curr->right]=true; } if(Keep_parent[curr] && !visted[Keep_parent[curr]]) { q.push(Keep_parent[curr]); visted[Keep_parent[curr]]=true; } } } vector ans; while(!q.empty()){ Node* temp = q.front(); q.pop(); ans.push_back(temp->data); } sort(ans.begin(),ans.end()); return ans; }

[@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; } };

[@tusharnain6652]

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

[@1234abcd-qt6uz]

for getting nodes at distance k visited map and array is not required only previous node is enough so that it doesn't call back (visited not required because this is not cyclic graph) for getting nodes at distance k CODE void dfsNodesAtDistanceK(TreeNode *node, TreeNode *pre, int k, unordered_map &parentMap, vector &ans) { if (k==0) { ans.push_back(node->val); return; } if (node->left && node->left!=pre) { dfsNodesAtDistanceK(node->left, node, k-1, parentMap, ans); } if (node->right && node->right!=pre) { dfsNodesAtDistanceK(node->right, node, k-1, parentMap, ans); } TreeNode *parent = parentMap[node]; if (parent!=NULL && parent!=pre) { dfsNodesAtDistanceK(parent, node, k-1, parentMap, ans); } return; } FULL CODE class Solution { private: void markParents(TreeNode *root, unordered_map &parentMap) { queue nodeQueue; nodeQueue.push(root); while (!nodeQueue.empty()) { TreeNode *node = nodeQueue.front(); nodeQueue.pop(); if (node->left) { parentMap[node->left] = node; nodeQueue.push(node->left); } if (node->right) { parentMap[node->right] = node; nodeQueue.push(node->right); } } return; } void dfsNodesAtDistanceK(TreeNode *node, TreeNode *pre, int k, unordered_map &parentMap, vector &ans) { if (k==0) { ans.push_back(node->val); return; } if (node->left && node->left!=pre) { dfsNodesAtDistanceK(node->left, node, k-1, parentMap, ans); } if (node->right && node->right!=pre) { dfsNodesAtDistanceK(node->right, node, k-1, parentMap, ans); } TreeNode *parent = parentMap[node]; if (parent!=NULL && parent!=pre) { dfsNodesAtDistanceK(parent, node, k-1, parentMap, ans); } return; } public: vector distanceK(TreeNode* root, TreeNode* target, int k) { unordered_map parentMap; markParents(root, parentMap); vector ans; dfsNodesAtDistanceK(target, NULL, k, parentMap, ans); return ans; } }; 863. All Nodes Distance K in Binary Tree

[@paritoshdadhich2954]

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

[@iWontFakeIt]

here, the problem in leetcode has constraints given that 0

[@harshjhunjhunuwala]

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

[@nagavedareddy5891]

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

[@apmotivationakashparmar722]

Thank you so much Striver !.

[@gubbalamalleswari5892]

Super explanation

[@adityaanand1912]

bhai kya gaaannnddddd faaaaadddd playlist hai maza aagaya

[@Vanshikataneja-r5w]

You are magic striver and you create magic

[@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

[@abhineetkapil]

Nice and easy explaination. !

[@ayushgautam8000]

GOD level explanation 🫡🫡

[@ishachauhan6477]

very well explained thankyou sir

[@vanshajtiwari1282]

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

[@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); } }

[@bhashkarbelwal4116]

you are amazing tutor #takeuForward bro

[@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!

[@harshmittal3128]

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

[@Abhishek-do8mp]

just one word - amazing

[@ransh-sahu]

the 2nd toughest quetion so far of this series

[@cinime]

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

[@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)

[@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.

[@sukhpreetsingh5200]

Amazing explanation😇

[@palakmantry]

Thank you for the explanation, really helpful

[@pranavsharma7479]

simple bfs traversal on the graph

[@Harsh-jc2bz]

its great question

[@051-avnee4]

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

[@jothsna]

Nice explanation☺

[@DeadPoolx1712]

UNDERSTOOD

[@YashKakrecha]

Wow nice explanation

[@sparshsharma6068]

Very well explained bhaiya! understood🔥

[@silicon9794]

Good explanation bro. Understood properly. Thankyou :)

[@AKASHKUMAR-li7li]

This can be easily solved using 1) root to node path 2) print all nodes k level down

[@mriduljain1981]

completed lecture 30 of free ka tree series.

[@rishabsharma5307]

please make videos on finding time complexity of finding complex questions

[@takeUforward]

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

[@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

[@radharamamohanakunnattur3035]

Understood!! awesome take

[@akhilkumarsingh5041]

hathi paar ki jai striver bahi ki

[@UECAshutoshKumar]

Thank you sir

[@subhankarpal2260]

Understood... thanks a lot.

[@rishabhshairy972]

Power of java streams no need to used for loops for final result. nodeQueue.stream().map(node -> node.val).collect(Collectors.toList());

[@ayushidalal5488]

Amazing explanation! Thankyou so much :)

[@suryakiran2970]

Great Explanation

[@mansisethi8127]

Dope question

[@prashantpandey2848]

we can avoid second loop using one more queue

[@arpitpachauri9189]

solved it without using queue and map :) class Solution { public: bool path(TreeNode* root,int x,vector&ans){ if(root==NULL) return false; ans.push_back(root); if(root->val==x) return true; if(path(root->left,x,ans) || path(root->right,x,ans)) return true; ans.pop_back(); return false; } void printkleveldown(TreeNode* root,int k,TreeNode* blocker,vector&v){ if(root==nullptr || kval); printkleveldown(root->left,k-1,blocker,v); printkleveldown(root->right,k-1,blocker,v); } vector distanceK(TreeNode* root, TreeNode* target, int k) { vectorans; vectorv; bool x=path(root,target->val,ans); reverse(ans.begin(),ans.end()); for(int i=0;i

[@NoName-hd7ok]

Store parents and do normal bfs using visited array

[@amisha2545]

🔥couldn’t have been better!

[@Learnprogramming-q7f]

Thank you Bhaiya

[@aakriti1]

Understood, Great explanation! 🤩

[@parthsalat]

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

[@himanshidafouty347]

Understood

[@ashwinnema06]

Your ending song is cool

[@AlokKumar-ld9qr]

amazing explanation. 😃😃😃

[@raven_claw887]

easy convert tree to graph

[@neerajgarg9096]

class Solution { private: void mapping(TreeNode* root, map&nodetoparent){ if(root==NULL){ return; } nodetoparent[root]=NULL; queueq; q.push(root); while(!q.empty()){ TreeNode* front=q.front(); q.pop(); if(front->left){ nodetoparent[front->left]=front; q.push(front->left); } if(front->right){ nodetoparent[front->right]=front; q.push(front->right); } } } public: vector distanceK(TreeNode* root, TreeNode* target, int k) { mapnodetoparent; mapping(root, nodetoparent); queueq; q.push(target); mapvisited; visited[target]=true; int d=0; while(d!=k){ int size=q.size(); for(int i=0; ileft && visited[front->left]==false){ visited[front->left]=true; q.push(front->left); } if(front->right && visited[front->right]==false){ visited[front->right]=true; q.push(front->right); } if(nodetoparent[front] && visited[nodetoparent[front]]==false){ visited[nodetoparent[front]]=true; q.push(nodetoparent[front]); } } d++; } vectorres; while(!q.empty()){ TreeNode* front = q.front(); res.push_back(front->val); q.pop(); } return res; }

[@g.upenderreddy]

we could use Set to keep track of visited nodes. public List distanceK(TreeNode root, TreeNode target, int k) { Map parentMap = markParentNodes(root); Set visited = new HashSet(); Deque queue = new LinkedList(); visited.add(target); queue.offer(target); int currentLevel = 0 , size = 0; while (!queue.isEmpty()) { if (currentLevel == k) break; currentLevel++; size = queue.size(); for (int i = 0; i< size; i++) { TreeNode node = queue.poll(); if (node.left != null && !visited.contains(node.left)) { queue.offer(node.left); visited.add(node.left); } if (node.right != null && !visited.contains(node.right)) { queue.offer(node.right); visited.add(node.right); } TreeNode parent = parentMap.get(node); if (parent != null && !visited.contains(parent)) { queue.offer(parent); visited.add(parent); } } } return queue.stream().map(node -> node.val).toList(); } public Map markParentNodes(TreeNode root) { Map parentMap = new HashMap(); // store Deque queue = new LinkedList(); queue.offer(root); while (!queue.isEmpty()) { TreeNode node = queue.poll(); if (node.left != null) { parentMap.put(node.left, node); queue.offer(node.left); } if (node.right != null) { parentMap.put(node.right, node); queue.offer(node.right); } } return parentMap; }