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?

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    /*
     * 采用前序遍历
     * 每一次递归将当前节点的val存入当前节点的list
     * 如果左右子节点的list不为空，则将返回后的左右子节点的list分别追加到当前节点的list中
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        List<Integer> left;
        List<Integer> right;
        
        if(root == null)
            return list;
        list.add(root.val);//中
        left = preorderTraversal(root.left);//左
        right = preorderTraversal(root.right);//右
         
        for(int i = 0; i < left.size(); i++){
            list.add(left.get(i));
        }
        for(int j = 0; j < right.size(); j++){
            list.add(right.get(j));
        }
        return list;
    }
}