<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>>ans;
    void DFS(TreeNode*root,int h){
        if(root==nullptr) return;
        if(h>=ans.size()) ans.emplace_back();
        ans[h].push_back(root->val);
        DFS(root->left,h+1);
        DFS(root->right,h+1);
    }
    vector<vector<int>> levelOrder(TreeNode* root) {
        DFS(root,0);
        return ans;
    }
};


        
    

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