## Leet code Question no: 102

**Companies asked this Question:**

*Company: no of times*

Amazon 2

Bloomberg 11

Uber 2

Apple 2

LinkedIn 25

Microsoft 13

Facebook 12

Oracle 7

Google 5

ServiceNow 3

Splunk 2

Adobe 2

Docusign 2

Accolite 2

TikTok 2

## Intuition

We aim to perform a level-order traversal of a binary tree, collecting nodes at each level into sublists.

## Approach

- Utilize a queue for level-order traversal, initializing an empty list
`WrapList`

to store results. - While the queue is not empty, dequeue nodes, collect their values into sublists, and enqueue their children.
- Add sublists to
`WrapList`

as levels are processed and return it as the result.

## Complexity

- Time complexity: O(n) - We visit each node once during the traversal.
- Space complexity: O(n) - The space usage grows with the number of nodes due to the queue and the result list.

## Code

```
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> WrapList = new LinkedList<>();
if(root == null)
return WrapList;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()){
int levelNum = queue.size();
List<Integer> subList = new LinkedList<>();
for(int i = 0 ; i < levelNum ; i++){
TreeNode node = queue.poll();
subList.add(node.val);
if(node.left != null) queue.offer(node.left);
if(node.right != null) queue.offer(node.right);
}
WrapList.add(subList);
}
return WrapList;
}
}
```

Happy coding,

shiva

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