**Inorder Traversal - Amazon Top Interview Questions**

### Problem Statement :

Given a binary tree root, return an inorder traversal of root as a list. Bonus: Can you do this iteratively? Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [1, null, [159, [80, null, null], null]] Output [1, 80, 159]

### Solution :

` ````
Solution in C++ :
vector<int> solve(Tree* root) { // Morris traversal - Time: O(N), Space: O(1)
vector<int> inorder;
while (root) {
if (root->left) {
Tree* predecessor = root->left;
while (predecessor->right && predecessor->right != root)
predecessor = predecessor->right;
if (predecessor->right) {
inorder.push_back(root->val);
root = root->right;
predecessor->right = nullptr;
} else {
predecessor->right = root;
root = root->left;
}
} else {
inorder.push_back(root->val);
root = root->right;
}
}
return inorder;
}
```

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
public int[] solve(Tree root) {
List<Integer> list = new ArrayList<>();
list = inOrder(root, new ArrayList<>());
int[] arr = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
arr[i] = list.get(i);
}
return arr;
}
public List<Integer> inOrder(Tree root, List<Integer> list) {
if (root == null)
return list;
Stack<Tree> st = new Stack<>();
Tree temp = root;
while (temp != null || st.size() > 0) {
while (temp != null) {
st.push(temp);
temp = temp.left;
}
temp = st.pop();
list.add(temp.val);
temp = temp.right;
}
return list;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root):
if not root:
return []
# iteratively
stack = [root.right, root.val, root.left]
traversal = []
while stack:
currNode = stack.pop()
if type(currNode) == int:
traversal.append(currNode)
elif currNode is None:
continue
else:
stack.extend([currNode.right, currNode.val, currNode.left])
return traversal
```

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