L10. iterative Inorder Traversal in Binary Tree | C++ | Java | Stack

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whatever you teach just goes straight into my head , thanks for making me feel not dumb.

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I bought a course from Coding Ninjas, I spent about 6000 on that and they have not even talked about iterative ways in the tree part. Thanks Striver bhai for making these high quality videos for free.

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haan sahi baat hai wo lavde paise khane bethe hain mc.

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mujhe offer mein 4100 ka milgya tha but still trees mein confidence nhi aaya. But ig maine trees thoda rush kia tha ,islie mujhe nhi aaya ache se samjh , course to acha hai vo no doubt, it all depends on the efforts ig in the end !!! Well i am glad you also have bought it,where are u now?

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Striver, Day 2 : completed this as well.

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Felt bad... your work deserves more likes ...those who watched it, please like this..so that it will improves sir's interest

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[@himanshu5549]

vector inorder(Node *root) { Node *curr = root; stack s; vector ans; while (true) { if (curr != NULL){ s.push(curr); curr = curr->left; } else { if (s.empty()) break; curr = s.top(); s.pop(); ans.push_back(curr->data); curr = curr->right; } } return ans; }

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Please never stop making such series. Loving it❤

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I don't know weather he is gifted, genius or really really hardworker. Anyways one thing is sure is a giver. Thanks Raj.

[@shivangisrivastava1158]

Very clearly explained! 👏 Just one thing, we can use root variable itself, instead of declaring 'node'. It still works! ( however no memory usage improvement on leetcode)

[@_theCorporateMonk]

Exactly!

[@Cool96267]

In that case we'll lose the root of tree, which we may require for further work.

[@saketmehta6697]

Modification of input ds is not at all recommended in interviews or real production env.

[@IBMX65]

@@_theCorporateMonk not a good practice brother

[@PrashantSingh-jy6zp]

for skip promotion go to 0:43 code explanation at 8:49

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most imp

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Thanks for making concepts easy to understand

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Completed 11/54(20% done) !!!

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Everything is very much clear to me thanks a lot bro

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I first like the video then starts watching 🤩....! Because I know your videos are always amazing. Thank you so much for such a amazing series.

[@adityajain5101]

A gentle feedback : The explanation is Great but here you could have given more light on the statement that how we are making iterative solution from recursion inorder solution discussed before ..... you have told that this is taken from recursive but with more explanation I think But Thankyou for the amazing series

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[@tanujamaurya8875]

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[@sukritinarang5016]

Recursive and Iterative both approaches are clear !Thanks

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please start string series also.It will be very helpful

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[@ganeshjaggineni4097]

NICE SUPER EXCELLENT MOTIVATED

[@shivalikagupta3433]

Thank you soo much !!! Recursive and Iterative clearing both the approaches. :))))

[@vineetjadhav1785]

Very Good Explaination❤❤🔥🔥

[@gaikwadpratik5872]

Thank you striver bhai. I think your way of teaching and delivering the concept is awesome. For me you are not lesser than GOD. I really don't have words to explain my feelings. @striver

[@prabhakaran5542]

Understood ❤

[@carefree_ladka]

Hey man , good work . However I wanna point out that you don't need to run your while loop infinitely and break it . We can simplify it as below: function inOrderTraversal(root) { const stack = []; const result = []; while (root || stack.length) { // Go as left as possible and push all left nodes to the stack while (root) { stack.push(root); root = root.left; } // Backtrack by popping from the stack root = stack.pop(); result.push(root.val); // Add the node value to the result // Now, visit the right subtree root = root.right; } return result; }

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[@samuelfrank1369]

UNDERSTOOD. THANKS A LOT.

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Thankyou Striver, Understood!

[@tps8470]

Easy approach

[@fantasyillusion]

Those who are looking for python solution # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]: inorder=[] node = root stack = [] while True: if node!=None: stack.append(node) node=node.left else: if len(stack)==0: break node=stack.pop() inorder.append(node.val) node=node.right return inorder

[@mannmehta4841]

Without infinite look condition, but O(N) extra space to track we have visited a child node or not - Code: void inordere_traversal_interative(Node* root) { stack st; st.push(root); unordered_map vis; while(!st.empty()) { Node* node = st.top(); if (node->left != NULL and !vis[node->left]) { st.push(node->left); continue; } coutright); continue; } } }

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thanks, bro for making such a wonderful series

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Thank you sir

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understood👍👍

[@RahulKumar-rk1tf]

Start: 0:43

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understooooooooood! Thanks again

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Liked. Cfbr will watch after some time

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[@aayushtheapple]

Alternative : ```cpp void inorder(Node* root) { if (root == NULL) return; stack st; Node* current = root; while (current != NULL || !st.empty()) { while (current != NULL) { st.push(current); current = current->left; } current = st.top(); st.pop(); cout data right; } } ```

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UNDERSTOOD;

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understood

[@armangaming8666]

really bohot amazing hai ye free ka tree series

[@parasgupta1391]

last mei root = node->right hoga naaki node = node->right

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[@vedanshsharma1156]

bhai phle hi stack empty hojayega 7 k left null par 7 jayega or right null par 1 jayega

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thank you very much please continue making such great videos

[@mriduljain1981]

completed lecture 10 of tree playlist.

[@Learnprogramming-q7f]

Thank you Bhaiya

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Wooowwww🎉

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Understood

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Thank you very much. You are a genius.

[@project_eth]

*understood*, day2

[@cinime]

Understood! Super cool explanation as always, thank you very much!!

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Thank you for this much effort and hard work.. ❤

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Thank u so much striver for such great Playlist

[@ujjwalsharmaiiitunacse3884]

dhanetawasi code :- vector inOrder(Node* root) { stack st; vector ans; if(root!=NULL) st.push(root); while(!st.empty()) { if(root==NULL) { st.pop(); if(st.empty()) break; root =st.top();st.pop(); ans.push_back(root->data); //right wali ko dalinge root=root->right; st.push(root); continue; } st.push(root->left); root=root->left; } return ans; }

[@pavankumar-lv4ne]

best coding youtuber ever

[@namansharma5128]

🚨🙋‍♂️📢Doubt- What is that while loop doing at line 19, we are not giving any condition, we have just written true? What does it signify?

[@KajalKumari-wy6ml]

It shows while( !stack.empty() || node != null)

[@chakshudeepmeharwal3547]

Line 26 has it

[@curs3m4rk]

finally understood this! much thanks to you

[@ideepakpandey]

thank u so much for this series.....

[@codemarshal655]

Did you know, you can do iterative in / pre / post order traversals using same code?? Checkout how I did it in just 20 lines, in my new video kzbin.info/www/bejne/bKjbf5ZunKidbqc

[@harshitjaiswal9439]

understood.

[@rupalikumari8829]

Understood : )

[@Mr.Indianwxvhehsy9191hghgx]

i like it

[@pankajsunal9819]

Understood .07/05/2024

[@uplsove]

This is for people who code in python def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]: stack = [] inorder = [] curr = root while True: if curr!=None: stack.append(curr) curr = curr.left else: if len(stack)==0: break curr = stack.pop() inorder.append(curr.val) curr = curr.right return inorder

[@lovetocode9266]

Is there any optimization while we are using iterative approach. As in iterative approach also we have time complexity of O(n) and aux.space O(n) . Isn't it same if we are using recurrive approach?

[@takeUforward]

Yeah its just for interviews.

[@Cool96267]

I don't think there can be any more optimization of time, as the name itself says 'traverse' we need to traverse whole of the n nodes.

[@bharadwajsaibandaru4323]

There's actually something called Morris traversal , only for inorder . You can check that out on GFG . It's a bit advanced I guess. It doesn't use recursion and stack even.

[@anuraj8979]

I understood concept...is any necessary to mugup code?

[@jaydalsaniya6986]

Press L four times at the starting of the video .

[@namansharma5128]

I would have to press it 8 times .......

[@_hulk748]

Understood sir❤🙏🙇‍♂

[@abhishekjakhmola1434]

amazing explaination

[@lavanyaprakashjampana933]

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

[@_AkashRuidas]

Great Explaination

[@jeetpatel7879]

while(true){ if(node!= NULL){ st.push(node); node = node->left; } else{ if(st.empty()== true) break; node = st.top(); ans.push_back(node->val); st.pop(); node= node->right; } } sir isme ye while(true) konsi condition cheak karta hei?? I mean when its going to be false?

[@patrickhudson5168]

With while(true), the loop will be running infinitely. But here, we have written a condition that when stack becomes empty, the loop will break.

[@jeetpatel7879]

@@patrickhudson5168 then how else condition will run??? You say is st.empty()==true then loop break then in loop how else condition will run??

[@shivangisrivastava1158]

@@jeetpatel7879 Note that break statement is inside an "if" condition, so it runs only when stack is empty, other times else block will get executed.

[@jeetpatel7879]

@@shivangisrivastava1158 when st.empty()==true then loop break then in loop how else condition will run? I checked that and also done dry run of code and the conclusion is when node!=NULL || !st.empty() this both condition is false then and only then loop will terminate

[@prabhagaikwad4849]

Below code is more cleaner, you may check once, public IList InorderTraversal(TreeNode root) { IList list = new List(); Stack stack = new Stack(); TreeNode node = root; while (node != null || stack.Count > 0) { if (node != null) { stack.Push(node); node = node.left; } else { node = stack.Pop(); list.Add(node.val); node = node.right; } } return list; }

[@rahulchandran6482]

Very helpful

[@sagarchauhan3187]

Hello Sir, Can you please complete the remaining videos of trees and graph of SDE sheet? please please upload!!🥺🥺

[@programmertik2046]

Here are the iterative traversals with explaination from me //Iterative traversals DFS You need to change only one line recursion simuated !! SC:O(2N) void InOrderIter(TreeNode *root){ unordered_mapstateMap; stacknodes; //push the root first nodes.push(root); while(!nodes.empty()){ TreeNode *currNode=nodes.top(); //if currNode is null simply pop it out if(currNode==NULL) { nodes.pop(); continue; } //if node is not present then initialize it in map //you may or may not do this as the map by default initializes //if key doesnt exist // if(stateMap.find(currNode)==stateMap.end()){ // stateMap[currNode]=0; // } //the nodes left side is not processed yet so process it if state of that node is 0 if(stateMap[currNode]==0){ nodes.push(currNode->leftPtr); } //the nodes left side is processed so process the root now and print its data it if state of that node is 1 else if(stateMap[currNode]==1){ cout

[@nagavedareddy5891]

Huge Respect...❤👏

[@sohamdachawar8354]

what si the while condition as in the code it is mentioned only true. but what is that true condition when should the loop stop working...I am unable to frame it in the condition.

[@RahulKumar-nd2sp]

Same doubt

[@II-ii2um]

What software are you using to draw these explanations man? Really nice colours you have there. Makes it light on the eyes.

[@shreyasnagabhushan4918]

understood sir

[@ashitasrivastava681]

UNDERSTOOD!

[@helloworldkhalid]

very nice sir thankyou so much

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Just Beautiful

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underStood!

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L 10 Done

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Keep up the good work

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[@rudranshvyas7539]

In c++ code what does "TreeNode *node = root;" is actually doing?? at 9:01

[@YashPandey_the_emperor]

It's just a Node iterator.

[@utkarshsharma6650]

understooood. thanks :)

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💚

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tysm striver bhai

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Understood 🖤

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Great content