/**
<p>给你一棵二叉树的根节点 <code>root</code> ，翻转这棵二叉树，并返回其根节点。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<p><img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/invert1-tree.jpg" style="height: 165px; width: 500px;" /></p>
<pre>
<strong>输入：</strong>root = [4,2,7,1,3,6,9]
<strong>输出：</strong>[4,7,2,9,6,3,1]
</pre>
<p><strong>示例 2：</strong></p>
<p><img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/invert2-tree.jpg" style="width: 500px; height: 120px;" /></p>
<pre>
<strong>输入：</strong>root = [2,1,3]
<strong>输出：</strong>[2,3,1]
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>root = []
<strong>输出：</strong>[]
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点数目范围在 <code>[0, 100]</code> 内</li>
<li><code>-100 <= Node.val <= 100</code></li>
</ul>
<div><div>Related Topics</div><div><li>树</li><li>深度优先搜索</li><li>广度优先搜索</li><li>二叉树</li></div></div><br><div><li>???? 1277</li><li>???? 0</li></div>
*/
//leetcode submit region begin(Prohibit modification and deletion)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root == null){
return null;
}
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
invertTree(root.left);
invertTree(root.right);
return root;
}
}
//leetcode submit region end(Prohibit modification and deletion)
不会，我可以学；落后，我可以追赶；跌倒，我可以站起来！
