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

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[@takeUforward]

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

[@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

[@gouravkumarshaw5467]

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

[@gouravkumarshaw5467]

Great!!

[@ridamsoni1762]

@@gouravkumarshaw5467 grt

[@Stefan_2117]

Gr8

[@roushankumar7684]

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

[@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

[@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 !!

[@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.

[@akankshasinha3352]

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

[@nilesh69420]

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

[@cinime]

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

[@stith_pragya]

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

[@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..

[@gandhijainamgunvantkumar6783]

Thank you so much for explaining very beautifully

[@roushankumar7684]

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

[@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.

[@prayagdesai4884]

Very clear and concise explaination..

[@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?

[@divakar4339]

Loved the concept. Thank You striver

[@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?

[@gay-bitchass-nigga3036]

@@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.

[@vishalgowrav]

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

[@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🙌

[@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.

[@iamsuryakant]

Awesome Explanation 👌

[@shreyasnagabhushan4918]

this was excellent bro, topic perfectly settled in my mind

[@sanchitsanyam7359]

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

[@prernagupta7729]

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

[@gouravkumarshaw5467]

great explanation thanks!!

[@roushankumar7684]

Thankyou bhaiya for lifting me up❤.

[@adrashsingh4109]

great explanation! Thank you

[@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; } };

[@sanjai_rs7]

Brilliant idea!

[@neelpatel122]

Brilliant Explanation

[@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!

[@lalitgaikwad7238]

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

[@iamnottech8918]

Superb explanation.

[@varnitgupta706]

amazing work

[@auroshisray9140]

Loving your explanation bhaiya😍

[@andrewcenteno3462]

Phenomenal explanation

[@iamnottech8918]

What an explanation !!

[@bhargavijethva7958]

Amazing!👏👏

[@vaalarivan_p]

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

[@user-tk2vg5jt3l]

Thank you Bhaiya

[@faizalam355]

Loved your content :)

[@bhaveshkumar6842]

Thank you, Striver!!!

[@siddharthmagadum16]

Your explanation is OP.

[@sriharshas3889]

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

[@26_jamkhandedattatray59]

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

[@arnedits6335]

grest explanantion.

[@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.

[@soubhikghosh6564]

Best solution ever!

[@KartikeyTT]

tysm sir

[@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

[@mohanseditz]

Good Explanation of code!

[@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

[@Shivi32590]

thank you

[@nagavedareddy5891]

Huge respect...❤👏

[@codewithree930]

Well explained

[@user-vo8wg3xr5w]

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

[@dharanit3918]

Excellent explanation bro❤❤

[@prathamchoudhary]

best explanation ever

[@clarencegomes6076]

Gazab! Thanks Striver

[@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

[@elements8630]

Awesome vid!!!

[@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

[@kuldeepmishra4718]

Maza aa gaya bhai. Aaise hi padhate raho

[@Anonymous-uj3jx]

Understood thanks :)

[@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🙃🙃

[@dipanshu-singh]

Love this vid. ❤

[@moinul6942]

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

[@varshiniramesh3752]

Yeah! I got it as well. It's the same code. But, why isn't mine working?

[@prajaktakeer487]

what is the change?

[@prajaktakeer487]

got it, just pass the vectors and map by reference

[@anonymous090]

Thank you Bhaiya❤

[@subhasishdas3011]

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

[@riyankadey102]

Amazing 🔥❤️

[@UECAshutoshKumar]

Thank you sir

[@jorgereis6943]

Thankyouuuuuuuuuuuuuuuuuuu, you're the man

[@alesblaze4745]

thanks mate!

[@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; } };

[@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

[@shivamchaudhary3646]

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

[@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

[@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.

[@PratikShah123]

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

[@Saurav_suman-pj8uk]

loving your video

[@ankitsingh9147]

Thankyou Bhaiya😀

[@deepaksarvepalli2344]

You are doing a great work

[@rishabhgupta9846]

understood ,nice explanation.What if duplicates values are present?

[@divyanshuyadav7278]

amazing🔥🔥🔥

[@harshadatrimukhe9147]

TLE for last test case(Leetcode)

[@sidakchuchra9643]

pass map by refrence\

[@AashimaKatoch]

@@sidakchuchra9643 it helped thankyou

[@avanishmaurya2034]

Nice

[@harshitjaiswal9439]

understood

[@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.

[@pirangitharun8736]

I like ur intro bro

[@bishalpandit4464]

Striver always op 🔥🔥

[@chitrrakarartwork9136]

thanks bro

[@vrandakansal5940]

🙌🙌

[@dreamyme543]

Understood:)

[@sujan_kumar_mitra]

Understood

[@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

[@symbol767]

I hate Math so much, figuring out how to calculate where these pointers should be is such a headache

[@kipa_chu]

Maths = CS lol

[@enigma368]

@@kipa_chu 🤣😂

[@jayeshbhanushali1745]

Please please please make DP series !!!