思路：

使用递归，归根结底还是左右节点互相倒，那么肯定需要一个temp节点在中间传递，最后就是递归，没什么说的        

代码：

/*** 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;reverse(root);return root;}public void reverse(TreeNode root) {if (root == null)return;TreeNode temp = root.left;root.left = root.right;root.right = temp;reverse(root.left);reverse(root.right);}
}