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

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

[takeUforward]

Hi, hope you are well.


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[pushkarraja]

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.

[deepaksarvepalli]

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

[PrashantSingh-jyzp]

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

[allinonep.]

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.

[shivangisrivastava]

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)

[aayushtheapple]

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

[ishwaripednekar]

Hello Striver, You explain code very well.
Thankyou for such kind of content.

[gaikwadpratik]

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

[shivalikagupta]

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

[sahilmathur]

Hello Striver, You explain code very well.
Thankyou for such kind of content.

[sanjana]

We love your series! And I can not thank you enough for your efforts!

[tanujamaurya]

Thank you Striver Bhaiya! for your efforts to make such great content !

[abhaytiwari]

thanks, bro for making such a wonderful series

[ujjwalsharmaiiitunacse]

dhanetawasi code :-
vector<int> inOrder(Node* root)
{
stack<Node*> st;
vector<int > 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;
}

[jambajuice]

striver is not a just a emotion . it is a name !

[himanshu]

vector<int> inorder(Node *root)
{
Node *curr = root;
stack<Node *> s;
vector<int> 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;
}

[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

[jasonbrody]

Liked. Cfbr will watch after some time

[nawabkhan]

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