<p>Given the <code>root</code> of a binary tree, invert the tree, and return <em>its root</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/invert1-tree.jpg" style="width: 500px; height: 165px;" />
<pre>
<strong>Input:</strong> root = [4,2,7,1,3,6,9]
<strong>Output:</strong> [4,7,2,9,6,3,1]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/invert2-tree.jpg" style="width: 500px; height: 120px;" />
<pre>
<strong>Input:</strong> root = [2,1,3]
<strong>Output:</strong> [2,3,1]
</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, 100]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</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 {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(root == NULL) return NULL;
        invertTree(root->left);
        invertTree(root->right);
        swap(root->left, root->right);
        return root;
    }
};
11600