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);
    }
}