144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:

Given binary tree 

{1,#,2,3}

,

1
\
2
/
3

return 

[1,2,3]

.

Note: Recursive solution is trivial, could you do it iteratively?
前序遍历用递归很容易写，用迭代需要用到堆栈。代码如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
List<Integer> res = new ArrayList<Integer>();
while (root != null) {
res.add(root.val);
if (root.right != null) {
stack.push(root.right);
}
root = root.left;
if (root == null && !stack.isEmpty()) {
root = stack.pop();
}
}
return res;
}
}