# LeetCode 102. Binary Tree Level Order Traversal

## Description

https://leetcode.com/problems/binary-tree-level-order-traversal/

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree `[3,9,20,null,null,15,7]`,

```    3
/ \
9  20
/  \
15   7
```

return its level order traversal as:

```[
[3],
[9,20],
[15,7]
]
```

## Explanation

Conduct a breadth-first search for level order traversal. Uses a queue to store the level nodes.

## Java Solution

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();

if (root == null) {
return result;
}

queue.offer(root);

while (!queue.isEmpty()) {
int levelSize = queue.size();
List<Integer> level = new ArrayList<>();

for (int i = 0; i < levelSize; i++) {
TreeNode node = queue.poll();

if (node.left != null) {
queue.offer(node.left);
}

if (node.right != null) {
queue.offer(node.right);
}
}

}

return result;
}
}``````

## 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 levelOrder(self, root: TreeNode) -> List[List[int]]:
results = []

if not root:
return results

queue = []
queue.append(root)

while queue:
level_size = len(queue)
level = []

for i in range(level_size):
node = queue.pop(0)
level.append(node.val)

if node.left:
queue.append(node.left)

if node.right:
queue.append(node.right)
results.append(level)

return results
``````
• Time complexity: O(N) since each node is processed exactly once.
• Space complexity: O(N) to keep the output structure that contains N node values.

## 2 Thoughts to “LeetCode 102. Binary Tree Level Order Traversal”

1. Priyanka Thakur says:

Your solution helped me understand Level Order Traversal. Thank you.

C# Solution :-
public class Solution {
public IList<IList> LevelOrder(TreeNode root) {
List<IList> AllList = new List<IList>();
var myqueue = new Queue();

if(root != null)
myqueue.Enqueue(root);

while(myqueue.Count>0)
{
int len = myqueue.Count;
var mylist = new List();

for(int i=0;i<len;i++)
{
var node = myqueue.Dequeue();