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?

思路1：trival的思路，recursive

    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> pre = new LinkedList<Integer>();
        preHelper(root,pre);
        return pre;
    }
    public void preHelper(TreeNode root, List<Integer> pre) {
        if(root==null) return;
        pre.add(root.val);
        preHelper(root.left,pre);
        preHelper(root.right,pre);
    }

思路2：用stack实现iteration'

public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {

        List<Integer> list = new ArrayList<Integer>();

        if (root==null){
            return list; 
        }

        Stack<TreeNode> st = new Stack<TreeNode>();
        st.push(root);

        while(!st.empty()){
            TreeNode node = st.pop();
            list.add(node.val);

            if (node.left!=null){
                st.push(node.left);            
            }

            if (node.right!=null){
                st.push(node.right);
            }


        }

        return list;
    }

}