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

Example 1:

```Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
```

Example 2:

```Input: root = [1]
Output: [[1]]
```

Example 3:

```Input: root = []
Output: []
```

Constraints:

• The number of nodes in the tree is in the range `[0, 2000]`.
• `-1000 <= Node.val <= 1000`

Idea:

Use BFS Template

Note: Don’t forget root === null case

Solution:

``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function(root) {
const res = [];
if (root === null) return res;

const queue = [];
queue.push(root);
while(queue.length !== 0) {
let size = queue.length;
let level = []; // store same level nodes
while(size--) {
let cur = queue.shift();
level.push(cur.val);
if (cur.left) queue.push(cur.left);
if (cur.right) queue.push(cur.right);
}
res.push(level.concat());
}
return res;
};``````