LeetCode Invert Binary Tree

Invert a binary tree.

4
   /   \
  2     7
 / \   / \
1   3 6   9

to

4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.思路分析：这是最近比较火的一个题目，因为一条推特的转播“Google HR：我们90%的工程师都用你写的软件，但是你竟然不会在白板上面反转一颗二叉树，所以滚吧”，各方看法不一，但看这个题目，的确是一个很简单的题目。考察最基本的递归和树操作。下面给出了递归实现和借助栈的迭代实现（非递归实现）。后一个版本的可扩展性更好，可以处理更大的树。实在是容易题，就是DFS遍历一遍树节点，把每个树节点的左右孩子互换就可以了。或许是那个ios开发牛人不屑于准备就去Google面试，结果被爆，与其说是能力问题，不如说是态度问题。AC Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode invertTree(TreeNode tn){
        /*if(tn == null) return null;
        TreeNode temp = tn.left;
        tn.left = tn.right;
        tn.right = temp;
        invertTree(tn.left);
        invertTree(tn.right);
        return tn;*/
        //0719
        if(tn == null) return null;
        Stack<TreeNode> tnStack = new Stack<TreeNode>();
        tnStack.push(tn);
        while(!tnStack.isEmpty()){
            TreeNode cur = tnStack.pop();
            TreeNode temp = cur.left;
            cur.left = cur.right;
            cur.right = temp;
            if(cur.left != null) tnStack.push(cur.left);
            if(cur.right != null) tnStack.push(cur.right);
        }
        return tn;
        //0726
    }
}