**Level Order Binary Tree to Linked List - Amazon Top Interview Questions**

### Problem Statement :

Given a binary tree root, convert it to a singly linked list using level order traversal. Constraints 1 ≤ n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [1, [2, null, null], [3, [4, null, null], [5, null, null]]] Output [1, 2, 3, 4, 5] Example 2 Input root = [1, [0, null, null], [2, null, null]] Output [1, 0, 2]

### Solution :

` ````
Solution in C++ :
void rec(Tree *root, int h, vector<LLNode *> &t) {
if (!root) return;
LLNode *tmp = new LLNode();
tmp->val = root->val;
if (t.size() >= h + 1) {
tmp->next = t[h];
t[h] = tmp;
} else {
tmp->next = nullptr;
t.push_back(tmp);
}
rec(root->right, h + 1, t);
rec(root->left, h + 1, t);
}
LLNode *solve(Tree *root) {
vector<LLNode *> t;
rec(root, 0, t);
LLNode *h, *dummy = new LLNode();
h = dummy;
for (LLNode *cur : t)
while (cur) {
dummy->next = cur;
cur = cur->next;
dummy = dummy->next;
}
return h->next;
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root):
Head = LL = LLNode(None)
d = deque()
d.append(root)
while d:
current = d.popleft()
LL.next = LLNode(current.val)
LL = LL.next
if current.left:
d.append(current.left)
if current.right:
d.append(current.right)
return Head.next
```

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