反復法を用いて二叉木を前順に遍歴する--Leetcodeシリーズ(七)

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?
My Answer

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List preorderTraversal(TreeNode root) {
        Stack stack = new Stack();
        List list = new ArrayList();
        if(root == null){
            return list;
        }
        stack.push(root);
        while(!stack.empty()){
            TreeNode top = stack.pop();
            list.add(top.val);
            if(top.right != null){
                stack.push(top.right);
            }
            if(top.left != null){
                stack.push(top.left);
            }
        }
        return list;
    }
}

テーマの出所:https://oj.leetcode.com/problems/binary-tree-preorder-traversal/