L34. Construct a Binary Tree from Preorder and Inorder Traversal | C++ | Java

[@takeUforward]

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

[@meme_eternity]

Relevel pe login otp nhi receive ho rahi..... fix it

[@SchrodingerMan]

for those who are getting TLE - pass the map in function with refernce ..................... it happens beacause when we pass map without refernce it start to copy all the value........ whereas if we pass the reference it only take base address................. so saving the time :-)

[@herculean6748]

Thanks man!!! I couldn't find why it was giving TLE

[@SchrodingerMan]

@@herculean6748 welcome 💪

[@AnkitKumarGupta03]

Thank you bro.

[@anshulkumarneekhara9377]

Thanks man

[@misterbean6674]

saved my life.

[@roushankumar7684]

Free me Aisa explanation koi nii dega. You are real life hero.

[@shubhamkumar-hx1fb]

bhut log mil jayenge ye tumhari fault hai....ager tum aankhen band karke khojoge to koi nhin dikhega

[@ransh-sahu]

@@shubhamkumar-hx1fb Searched the whole youtube being comfortable and understanding it very well . The consistency striver followed is unmatched and its true none is better than striver if u r targeting fang or higher . There is always a shortage of content or room for subscription. Praise striver rather than being cool lgunga ase bolunga

[@samyakjain7422]

coding ninja ka course liya tha usme iski explanation smjh nhi aayi lekin apki video dekhi aur 1 shot me smjh aagai . thanks striver . u teach like u r my friend and not a teacher which i like most bout ur videos !!

[@gouravkumarshaw5467]

(a) Inorder (Left, Root, Right) : (b) Preorder (Root, Left, Right) : (c) Postorder (Left, Right, Root) :

[@gouravkumarshaw5467]

Great!!

[@ridamsoni1762]

@@gouravkumarshaw5467 grt

[@Stefan_2117]

Gr8

[@satyamraj2779]

grate.

[@manaskhare412]

It took a while to understand this tricky problem, but now it's completely clear. Thank you striver, kudos for such amazing content.

[@alexrcrew1975]

haha nice going at present whichtopic youaredoing

[@cur92lyone]

I spent two days trying to understand this Algo. Finally, your video made me understand it. Thanks. :)

[@rushidesai2836]

I can understand. Two of the most important quesitons - Tree from Preorder and Inorder, and tree from inorder and postorder.

[@gouravchaki7990]

Out of all these coding I would just provide method to prepare Biriyani!! Step 1 Prepare saffron-kewra water and chop veggies To make a delightful chicken biryani dish, firstly soak saffron in water to prepare saffron water (one tsp saffron can be soaked in 1/4 cup water). Next, mix kewra drops in water and mix well to make kewra water. Set them aside for later usage. Now, chop the onion and coriander leaves and keep them aside. Step 2 Saute the onions Meanwhile, heat olive oil in a deep bottomed pan. Once the oil is hot enough, add cumin seeds, bay leaf, green cardamom, black cardamom, cloves in it, and saute for about a minute. Then, add chopped onion to it and saute until pink. Now, add chicken into it with slit green chillies, turmeric, salt to taste, ginger-garlic paste, red chilli powder and green chilli paste. Mix well all the spices and cook for 2-3 minutes. Then, add hung curd into it and give a mix. (Make sure the chicken is washed properly and patted dry before adding it to the dish) Step 3 Cook biryani on low heat for 5-6 minutes Turn the flame to medium again and add garam masala in it along with ginger julienned, coriander and mint leaves. Add kewra water, rose water and 1 tsp saffron water in it. Cook till the chicken is tender. Then add 1 cup cooked rice and spread evenly. Then add the remaining saffron water and pour ghee over it. You can now cook the dish without the lid or cover it with a lid to give a dum-effect due to the steam formation. Step 4 Serve hot chicken biryani with your favourite chutney or raita Cook for 15-20 minutes with a closed lid and garnish with 1 tbsp fried onions and coriander leaves. Serve hot chicken biryani with raita of your choice. Enjoy!

[@NitishGautam-g1j]

Bro we need more teachers like you 😢😢😢 who actually can think from perspective of a beginner and explain step by step

[@anubhav452]

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ /* JUST FINDING THE POSITION OF THE FIRST ELEMENT IN THE PREORDER VECTOR IN THE INORDER VECTOR THEN USING THE POSITION OF THAT ELEMENT TO FIND THE LEFT SUBTREE AND THE RIGHT SUBTREE */ class Solution { public: int find_pos(vectorv,int x){ for(int i=0;ileft=buildTree(l1,l2); root->right=buildTree(r1,r2); return root; } };

[@roushankumar7684]

You have changed the teachers image as spreading skill without taking any money❤

[@ss-pz4zh]

THE BEST TREE series even though i have taken a paid course...still i am coming here to understand ...>God Bless u TUF

[@gautamkrishnan9982]

Keep Going Striver!! I am following your SDE Sheet to revise on the DSA concepts. Thank you so much for all your content.

[@Cool96267]

Which programming language you are using?

[@mynk_rjpt]

@@Cool96267 c++

[@9-teen77]

did you complete it?

[@Parth.Deshpande]

For interviewbit solution , pass the map by reference TreeNode* buildT(vector &A,int prestart,int preend,vector &B,int instart,int inend , map& mp){ if(prestart>preend || instart>inend){ return NULL; } TreeNode* root = new TreeNode(A[prestart]); int inroot = mp[root->val]; int leftnums = inroot - instart; root->left = buildT(A,prestart+1,prestart+leftnums,B,instart,inroot-1,mp); root->right = buildT(A,prestart+leftnums+1,preend,B,inroot+1,inend,mp); return root; } TreeNode* Solution::buildTree(vector &A, vector &B) { map mp; for(int i=0;i

[@priyanshurai460]

Can you please tell me why the time limit exceeds when the map is passed by value?

[@Star_Gazer_42]

@@priyanshurai460 coz everytime the function gets called it creates a new map with same name at another memory location with same values. So these operations take time and space which is alot, thats why TLE.

[@natanshisharma8828]

If you are confused how the they are storing addresses, its during backtracking that the nodes addresses are being stored not during creation of the nodes.

[@ShreyaJain-1808]

Here is the code to deal with duplicate elements we use a queue so that the element that occurs later in the in-order get pushed at the front and keep popping the index that are processed, we now get a correct tree formed according to the preorder traversal class Solution{ public: Node* solve(int in[],int in_start,int in_end,int pre[],int pre_start,int pre_end,map &pos) { if(pre_start>pre_end|| in_start>in_end) { return NULL; } int inroot=pos[pre[pre_start]].front(); pos[pre[pre_start]].pop(); int elem=inroot-in_start; Node* root=new Node(pre[pre_start]); root->left=solve(in,in_start,inroot-1,pre,pre_start+1,pre_start+elem,pos); root->right=solve(in,inroot+1,in_end,pre,pre_start+elem+1,pre_end,pos); return root; } Node* buildTree(int in[],int pre[], int n) { // Code here if(n==0) return NULL; map pos; for(int i=0;i

[@meetmandhane]

Will the time complexity of this approach remain same i.e. O(N)?

[@ayushgautam169]

yeah exactly what i did i took a list and removed from front, like we can take a queue

[@i-am-darshil]

Is a unique tree even possible in case of duplicate elements? 2 trees can be formed with the same inorder and preorder Inorder - 1 2 2 3 2 Preorder - 2 2 1 2 3

[@tanishq2766]

I guess we don’t need a map, if we can split them as we can always split into two parts.

[@trojanhorse8278]

@@i-am-darshil Correct, You can't always construct a unique binary tree when the keys are repeated. EX is Pre : 10 10 10 In : 10 10 10 , now based on your choice of root in in-order you will get different tree structures. Hence only if the interviewer wants to check whether you know this fact or not he might ask this question but once you explain to him that we can't always create a unique tree without defining the rule to choose the root from in-order, he mostly will get satisfied and won't ask you to actually write a code.

[@akankshasinha3352]

Thankyou striver 🥺 you’re a gem that all of us have got !!

[@karthik-varma-1579]

Java Code:- class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { HashMap hm = new HashMap(); int n = inorder.length; for(int i=0;ipreEnd || inStart>inEnd) return null; TreeNode root = new TreeNode(preorder[preStart]); int hMap = hm.get(root.val); int numsLeft = hMap-inStart; root.left = builder(inorder,inStart,hMap-1,preorder,preStart+1,preStart+numsLeft,hm); root.right = builder(inorder,hMap+1,inEnd,preorder,preStart+numsLeft+1,preEnd,hm); return root; } }

[@shikharsrivastava7312]

simplified version of code explained above /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: TreeNode* constructTree(int i,int j,int &ind,vector &preorder,vector &inorder,unordered_map &mp) { if(i>j) { return NULL; } int rootVal=preorder[ind]; ind+=1; TreeNode *root=new TreeNode(rootVal); int in=mp[rootVal]; root->left=constructTree(i,in-1,ind,preorder,inorder,mp); root->right=constructTree(in+1,j,ind,preorder,inorder,mp); return root; } TreeNode* buildTree(vector& preorder, vector& inorder) { unordered_map mp; for(int i=0;i

[@AmitSharma-nv2oj]

can you explain how i>j condition works

[@stith_pragya]

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

[@nilesh69420]

Great explanation as always. The last dry run just before showing code was very very helpful.

[@AppaniMadhavi]

Thanks alot bhai!!. I didn't understood first but in other problems I need to use this, so I visited once again and now its clear cut for me thanks alot..

[@lalitgaikwad7238]

You're The Real Life G.O.A.T i.e Greatest of All Times

[@sammy_yo5967]

C++ code to run on Leetcode: - class Solution { public: TreeNode* buildTree(vector& preorder,int preStart,int preEnd,vector& inorder,int inStart,int inEnd, map& inMp) { if(preStart > preEnd || inStart > inEnd) return NULL; TreeNode* root = new TreeNode(preorder[preStart]); int inRoot = inMp[root->val]; int numsLeft = inRoot - inStart; root ->left = buildTree(preorder,preStart+1,preStart+numsLeft,inorder,inStart,inRoot - 1,inMp); root -> right = buildTree(preorder,preStart+numsLeft+1,preEnd,inorder,inRoot+1,inEnd,inMp); return root; } TreeNode* buildTree(vector& preorder, vector& inorder) { map inMp; for(int i = 0; i < inorder.size(); i++) { inMp[inorder[i]] = i; } TreeNode* root = buildTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1,inMp); return root; } };

[@agnijeetsharma]

here is simple version of code class Solution { public: TreeNode* build(vector& preorder, vector& inorder, int& preStart, int inStart, int inEnd,unordered_map&mp) { if (inStart > inEnd) return nullptr; TreeNode* curr = new TreeNode(preorder[preStart]); int pos=mp[preorder[preStart]]; preStart++; // Move to the next root in the preorder sequence curr->left = build(preorder, inorder, preStart, inStart, pos - 1,mp); curr->right = build(preorder, inorder, preStart, pos + 1, inEnd,mp); return curr; } TreeNode* buildTree(vector& preorder, vector& inorder) { unordered_mapmp; for(int i=0;i

[@sourabhsharma2746]

This is toh another level explanation bhai. 👌👌 how can someone be so clear in his explanation.. btw amazing thanks bhai. love from punjab

[@rushidesai2836]

Yeh baat pe bhangdaa ho jae! :)

[@jotsinghbindra8317]

chki rkh kam bai🙌

[@sriharshas3889]

Superb Explanation ! Here in the code we should pass the map as reference otherwise we will get the memory limit exceeded.

[@vaalarivan_p]

3:30 - 3:35 CW : 12:40 - 17:01

[@akashverma5756]

This is one of insanely tricky problem of Binary Tree.

[@ankushsingh3392]

Vro dus video dekhi but nhi hua apki video ka sirf first two minutes ke logic ko dekh kar samjh gya and done it🙃🙃

[@sri199]

For duplicates value try dict of list to store all index of the element and pop the first one res = [] mmap = defaultdict(list) n = len(preorder) def build(preStart,preEnd,inStart,inEnd): if preStart > preEnd or inStart > inEnd: return None root = TreeNode(preorder[preStart]) orders = mmap[root.val] index = orders.pop(0) inRoot = index numsLeft = inRoot - inStart root.left = build(preStart+1,preStart+numsLeft,inStart,inRoot-1) root.right = build(preStart+numsLeft+1,preEnd,inRoot + 1,inEnd) return root for i in range(len(inorder)): mmap[inorder[i]].append(i) return build(0,n-1,0,n-1)

[@sarthak._v_]

Best Lecture

[@26_jamkhandedattatray59]

doing dry run is little bit difficult pr when i understood 😌 toh bas mazza agaya😀. Thnx bro for such great video 😊

[@vinayrohitreddypadala2705]

Hey Striver, this algorithm is working for unique values but not for the repeated ones. Example:- inorder = [7 ,3 ,11 ,1 ,17 ,3 ,18] postorder = [1, 3, 7, 11, 3, 17, 18] And In this scenario i dont think inMap is kind of working.

[@KandulaAsrith]

someone please reply how to solve if the tree contain duplicate elements...

[@TonyStark-ct8xd]

This code doesn't work for duplicate values, gfg has updated their testcases.

[@rushidesai2836]

Khud se try karna bhai, he is already putting out so much content for us.

[@apmotivationakashparmar722]

Thank you So much Striver !

[@KeigoEdits]

I am shocked, that i came up with the solution to this problem in my first attempt, though runtime was low as i was searching index of root in inorder using linear search(though dividing the space in half after each recursive call) but its still O(n). Later saw in this video, that we can use hashmap for this purpose. Though indeed I couldn't have done without the intuition I got in earlier video of striver explaining unique binary question.

[@iamnottech8918]

Superb explanation.

[@sanchitsanyam7359]

Thanks allot Bhaiya ,your videos awesome , I love your videos.

[@roushankumar7684]

Thankyou bhaiya for lifting me up❤.

[@gandhijainamgunvantkumar6783]

Thank you so much for explaining very beautifully

[@varnitgupta706]

amazing work

[@nextnotification9857]

wrote the code in first attempt😀, i have made the map global to the class and thus no need to pass it again and again

[@iamnottech8918]

What an explanation !!

[@MJBZG]

understood striver, much love

[@CharitraAgarwal]

TreeNode *getNode(vector&pre, vector& ino, int s, int e, int &i) { if(s > e) return NULL; // no more elements in inorder range int el = pre[i++]; // get current element in preorder TreeNode *node = new TreeNode(el); // create a node if(s == e) return node; // if its the only node in inorder range // search for this node in the inorder arrangement int ind = s; while(ino[ind] != el) ind++; // if there are some elements in the left of the inorder index, // this means, we have some elements in the left subtree if(ind > s) node->left = getNode(pre, ino, s, ind-1, i); // remaining elements are in right subtree node->right = getNode(pre, ino, ind+1, e, i); return node; // return this node } TreeNode* buildTree(vector& preorder, vector& inorder) { int i = 0; // index for preorder array // taking preorder index as reference, as it gets updated as soon as // elements gets used return getNode(preorder, inorder, 0, inorder.size()-1, i); }

[@vishalgowrav]

What an explanation!!🥺❤ Take a bow striver🙇‍♂️

[@sanjai_rs7]

Brilliant idea!

[@andrewcenteno3462]

Phenomenal explanation

[@kuldeepmishra4718]

Maza aa gaya bhai. Aaise hi padhate raho

[@adrashsingh4109]

great explanation! Thank you

[@arnedits6335]

grest explanantion.

[@siddharthmagadum16]

Your explanation is OP.

[@Learnprogramming-q7f]

Thank you Bhaiya

[@harshadatrimukhe9147]

TLE for last test case(Leetcode)

[@AashimaKatoch]

@Sidak Chuchra it helped thankyou

[@cinime]

Understood! So awesome explanation as always, thank you very much!!

[@subhasishdas3011]

in base case no need to check both in and pre any 1 check is enough

[@moinul6942]

Do check out the code link provided in the description to avoid TLE on leetcode.

[@codding32world50]

For TLE on leetcode just declare map public don't pass it in the function

[@adarshpandey622]

Bro you saved my whole day

[@saiprashanth2646]

Thank u so much bro.

[@tejuschaturvedi6234]

you can also pass the map as reference when passing the map in function .

[@prashukjain8519]

saved my time

[@soubhikghosh6564]

Best solution ever!

[@bhargavijethva7958]

Amazing!👏👏

[@PratikShah123]

Wonderful !! Btw, what if Tree contains duplicate values, where finding index out of InOrder will be a challenge..!

[@KartikeyTT]

tysm sir

[@Shivi32590]

thank you

[@Lalit_Shirsath]

can we use unordered map to save time complexity ? map --> O( log n ) unordered_map --> O(1)

[@takeUforward]

Unordered map’s worst case is still o(n) 😉

[@Lalit_Shirsath]

@@takeUforward ha bhaiya 😁🤗

[@shivammishrasvnit5456]

@@takeUforward bhyaa still in leetcode tle coming what to do😞😞 ?

[@shivammishrasvnit5456]

no i got accepted after putting map by reference, thanks for it you explained soo well!

[@hiddenguy7174]

@@shivammishrasvnit5456 try passing map by reference

[@GoluYadav-cg3kx]

Hi striver love from iit r. I just wanna clarify that if case the in order traversal have duplicate values then what should be our approach

[@samarthvarshney512]

just use a unordered_map> mp

[@prathamchoudhary]

best explanation ever

[@UECAshutoshKumar]

Thank you sir

[@prernagupta7729]

You are doing a great job bro...thanks a lot.

[@prayagdesai4884]

Very clear and concise explaination..

[@divakar4339]

Loved the concept. Thank You striver

[@shivamchaudhary3646]

Very well explained... but at leetcode it will give memory limit exceed for a particular testcase make map globally !

[@sparshsharma6068]

Understood! but, Bhaiya There is a more optimal version available in Leetcode discussion.

[@takeUforward]

I don’t think so.. can you please paste it here.

[@sparshsharma6068]

@@takeUforward Here's the java code bhaiya private TreeNode constructTree(TreeNode root, int[] idx, int stop,int[] preorder, int[] inorder){ if(idx[0] >= preorder.length) return root; if(inorder[idx[1]] == stop){ idx[1]++; return root; } root = new TreeNode(preorder[idx[0]++]); root.left = constructTree(null, idx, root.val, preorder, inorder); root.right = constructTree(null, idx, stop, preorder,inorder); return root; } public TreeNode buildTree(int[] preorder, int[] inorder) { int[] idx = {0,0}; //{preorder_idx, inorder_idx} return constructTree(null, idx, Integer.MAX_VALUE, preorder, inorder); } Time O(n), Space O(n) (recursive) I kept on dry running it and it then seemed logical, but, I still don't know the intuition behind it. It will be very helpful, If can you make a LC post and explain the intuition behind this. link to discussion : leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/discuss/34543/Simple-O(n)-without-map Look in the comments, you will find C++ or java code.

[@adithyaas7735]

202/203 test case timelimit exceeded to fix this just put ''&map" on function arguments

[@santanu29]

Thank you. I was stuck there. Quickly can you mention why it gives tle for not using "&"

[@adithyaas7735]

@@santanu29'&' means we are passing the map as reference (only the address of it) In pass by reference, no new copy of the variable is made, so overhead of copying is saved

[@anonymous090]

Thank you Bhaiya❤

[@viditaagrawal4169]

Nice explanation !! For duplicate values instead of using map we can create individual function findIndex from inorder array using visited array , which won't repeat index , my code passed on gfg

[@nikhilsachan_cse7144]

In question, it is written that no duplicate values are allowed, otherwise it will be impossible to create tree from traversals in some cases.

[@fine2981]

can u pls share ur code.??

[@viditaagrawal4169]

@@fine2981 int findIndex(int in[],int ele,int n,vector &vis){ for(int i=0;ipreEnd || inStart>inEnd) return NULL; Node* root = new Node(pre[preStart]); int index = findIndex(in,root->data,n,vis); int inleftEnd = index-1; int preleftEnd = preStart+index-inStart; root->left = solve(in,pre,inStart,inleftEnd,preStart+1,preleftEnd,vis,n); int inrightSt = index+1; int prerightSt = preleftEnd+1; root->right = solve(in,pre,inrightSt,inEnd,prerightSt,preEnd,vis,n); return root; } public: Node* buildTree(int in[],int pre[], int n) { vector vis(n,0); return solve(in,pre,0,n-1,0,n-1,vis,n); }

[@viditaagrawal4169]

@@nikhilsachan_cse7144 Its on leetcode that duplicate isn't there but on gfg its mentioned in question that duplicates are also present

[@fine2981]

/*Complete the code here. Node is as follows: struct Node { int data; Node* left; Node* right; }; */ class Solution{ public: Node* tree(int in[],int pre[], int &preindex,int ins,int ine,int n){ if(ins>ine || preindex>=n){ return NULL; } int element=pre[preindex++]; Node* root=new Node(element); int inorderindex=0; for(int i=0;idata==in[i]){ inorderindex=i; break; } } root->left=tree(in,pre,preindex,ins,inorderindex-1,n); root->right=tree(in,pre,preindex,inorderindex+1,ine,n); return root; } Node* buildTree(int in[],int pre[],int n) { int preindex=0; Node* ans=tree(in,pre,preindex,0,n-1,n); return ans; } }; can you tell me whats wrong with this code? its giving me wrong ans after 54 test cases.. this was the test case that it did not passed. 7 7 3 11 1 17 3 18 1 3 7 11 3 17 18

[@bhaveshkumar6842]

Thank you, Striver!!!

[@dsa.aditya]

IN GFG THE CODE MIGHT GET SEGMENTATION FAULT SO PLEASE USE QUEUE DATA STRUCTURE TO STORE THE MULTIPLE INDEX OF A SAME ELEMENT. REALLY Thank You so much Striver Bhaiya🙏 HERE IS THE CODE ------ Node* ans(int pre[],int preStart,int preEnd,int in[],int inStart,int inEnd,unordered_map& mp) { if (preStart > preEnd || inStart > inEnd) { return NULL; } Node* root=new Node(pre[preStart]); int inroot=mp[root->data].front(); mp[root->data].pop(); int left=inroot-inStart; root->left=ans(pre,preStart+1,preStart+left,in,inStart,inroot-1,mp); root->right=ans(pre,preStart+left+1,preEnd,in,inroot+1,inEnd,mp); return root; } Node* buildTree(int in[], int pre[], int n) { unordered_map mp; for (int i = 0; i < n; i++) { mp[in[i]].push(i); } Node* root=ans(pre,0,n-1,in,0,n-1,mp); return root; }

[@relaxingnaturalvibrations1171]

thanks bro

[@mohanseditz]

Good Explanation of code!

[@vvsish8803]

Python Solution class Solution: def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]: inMap={} for i in range(len(inorder)): inMap[inorder[i]]=i return self.helper(preorder,0,len(preorder)-1,inorder,0,len(inorder)-1,inMap) def helper(self,preorder,preStart,preEnd,inorder,inStart,inEnd,inMap): if preStart>preEnd: return None elif preStart==preEnd: return TreeNode(preorder[preStart]) else: rootVal=preorder[preStart] inRoot=inMap[rootVal] numsLeft=inRoot-inStart root=TreeNode(rootVal) root.left=self.helper(preorder,preStart+1,preStart+numsLeft,inorder,inStart,inRoot-1,inMap) root.right=self.helper(preorder,preStart+numsLeft+1,preEnd,inorder,inRoot+1,inEnd,inMap) return root

[@shreyasnagabhushan4918]

this was excellent bro, topic perfectly settled in my mind

[@dharanit3918]

Excellent explanation bro❤❤

[@isheep9025]

self note:t inleft=no of elements left on the left of root index. inroot=index of root in inorder inlefr- used for indexing pre/post; inroot-used for indexing inorder

[@003_sambit5]

Thanks a lot Striver for providing such amazing content to us for free of cost...respect for your Hard Work. Just one doubt , not sure if leetcode has updated its tcs coz now this algo is giving TLE at 202nd tc out of 203 tcs...

[@013_aman]

Take map as by reference it will work

[@SahilSharma-to8nd]

@@013_aman thanks

[@gouravkumarshaw5467]

great explanation thanks!!

[@neelpatel122]

Brilliant Explanation

[@VishakhavelShanmuganathan]

thanks!

[@avanishmaurya2034]

Nice

[@jayeshbhanushali1745]

Please please please make DP series !!!

[@DeadPoolx1712]

UNDERSTOOD;

[@auroshisray9140]

Loving your explanation bhaiya😍

[@Saurav_suman-pj8uk]

loving your video

[@supratimbhattacharjee5324]

class Solution { public: TreeNode* create(vector& preorder, vector& inorder, unordered_map& hash,int ps,int pe,int is,int ie) { if(ps>pe || is>ie) return nullptr; int rootIndxInInorder=hash[preorder[ps]]; int lps=ps+1; int lpe=ps+rootIndxInInorder-is; int rps=lpe+1; int rpe=pe; int lis=is; int lie=rootIndxInInorder-1; int ris=rootIndxInInorder+1; int rie=ie; TreeNode* root=new TreeNode(preorder[ps]); TreeNode* left=create(preorder,inorder,hash,lps,lpe,lis,lie); TreeNode* right=create(preorder,inorder,hash,rps,rpe,ris,rie); root->left=left; root->right=right; return root; } TreeNode* buildTree(vector& preorder, vector& inorder) { int n=preorder.size(); unordered_map hash; for(int i=0;i

[@ankitsingh9147]

Thankyou Bhaiya😀

[@akashbhoyar7403]

leetcode added some crazy test case......to which striver sir's code is showing TLE. If anyone knows abt -> How to deal with TLE....Please Comment !

[@als_venky7057]

just send your map as a reference instead of sending it normally. And see the magic, boooom💥💥

[@alesblaze4745]

thanks mate!

[@clarencegomes6076]

Gazab! Thanks Striver

[@achalsayee2534]

note :In case of duplicates map won't work.because it overwrites old index with latest index. so do linear search within the bounds in inStart and inEnd.

[@vijaynag7723]

understood

[@dipanshu-singh]

Love this vid. ❤