**Level Order Traversal - Amazon Top Interview Questions**

### Problem Statement :

Given a binary tree root return a level order traversal of the node values. Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [0, [5, null, null], [9, [1, [4, null, null], [2, null, null]], [3, null, null]]] Output [0, 5, 9, 1, 3, 4, 2] Example 2 Input root = [0, [1, [2, [3, null, null], null], null], null] Output [0, 1, 2, 3] Example 3 Input root = [0, null, [1, null, [2, null, [3, null, null]]]] Output [0, 1, 2, 3]

### Solution :

` ````
Solution in C++ :
vector<int> solve(Tree* root) {
vector<int> res;
if (root == nullptr) return {};
queue<Tree*> q;
q.push(root);
while (!q.empty()) {
Tree* n = q.front();
q.pop();
res.push_back(n->val);
if (n->left) q.push(n->left);
if (n->right) q.push(n->right);
}
return res;
}
```

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
public int[] solve(Tree root) {
if (root == null)
return new int[0];
ArrayList<Integer> list = new ArrayList<>();
Queue<Tree> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
Tree pop = queue.poll();
list.add(pop.val);
if (pop.left != null) {
queue.offer(pop.left);
}
if (pop.right != null) {
queue.offer(pop.right);
}
}
return convertIntegers(list);
}
public int[] convertIntegers(List<Integer> integers) {
int[] ret = new int[integers.size()];
Iterator<Integer> iterator = integers.iterator();
for (int i = 0; i < ret.length; i++) {
ret[i] = iterator.next().intValue();
}
return ret;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root):
q = [root]
ans = []
for node in q:
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
ans.append(node.val)
return ans
# ans = []
# if root:
# if a==True:
# ans.append(root.val)
# if root.left:
# ans.append(root.left.val)
# if root.right:
# ans.append(root.right.val)
# ans += self.solve(root.left,False)
# ans += self.solve(root.right,False)
# return ans
```

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