<p>Given the <code>root</code> of a binary tree, return <em>the level order traversal of its nodes&#39; values</em>. (i.e., from left to right, level by level).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [3,9,20,null,null,15,7]
<strong>Output:</strong> [[3],[9,20],[15,7]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> [[1]]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = []
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 2000]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> order;
        if (!root) return order;

        queue<TreeNode*> q;
        q.push(root);
        int size;
        while (!q.empty()) {
             size = q.size();

            vector<int> currentLevel;
            for(int i = 0; i < size; ++i){
                TreeNode* n = q.front();
                q.pop();
                currentLevel.push_back(n->val);
                if(n->left) q.push(n->left);
                if(n->right) q.push(n->right);
                n->left = n->right = nullptr;
            }
            order.emplace_back(currentLevel);
        }

        return order;
    }
};

12000