Lintcode 175. 翻转二叉树

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递归那么好为什么不用递归啊...我才不会被你骗...（其实是因为用惯了递归啰嗦的循环反倒不会写了...o(╯□╰)o）

AC代码：

/**
* Definition of TreeNode:
* public class TreeNode {
*     public int val;
*     public TreeNode left, right;
*     public TreeNode(int val) {
*         this.val = val;
*         this.left = this.right = null;
*     }
* }
*/
public class Solution {
/**
* @param root: a TreeNode, the root of the binary tree
* @return: nothing
*/
public void invertBinaryTree(TreeNode root) {
if(root==null) return ;
TreeNode t=root.left;
root.left=root.right;
root.right=t;
invertBinaryTree(root.left);
invertBinaryTree(root.right);
}
}

题目来源： http://www.lintcode.com/zh-cn/problem/invert-binary-tree/