<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>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    int getHeight(TreeNode* root){
        if (!root){
            return 0;
        }
        return max(getHeight(root -> left), getHeight(root -> right)) + 1;
    }

    void bfs (TreeNode* root, int level, vector<int>& vec) {
        if (!root || level <= 0) {
            return;
        }
        if (level == 1) {
            vec.push_back(root -> val);
        } else {
            bfs(root -> left, level - 1, vec);
            bfs(root -> right, level - 1, vec);
        }
    }
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        int h = getHeight(root);
        vector<vector<int>> vec (h);
        for (int i = 1; i <= h; ++i){
            bfs(root, i, vec[i - 1]);
        }
        return vec;
    }
};
14500