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 binary tree
 * 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> list = new ArrayList<Integer>();
        while(root != null){
            while(root.left != null){
                stack.push(root);
                list.add(root.val);
                root = root.left;
            }
            if(root.right != null){
                list.add(root.val);
                root = root.right;
            }else{//叶子节点
                list.add(root.val);
                if(stack.empty()){//最后一个叶子
                    break;
                }
                while(!stack.empty()){//有祖先节点
                    TreeNode node = stack.pop();//返回栈顶元素
                    if(node.right != null){
                        root = node.right;
                        break;
                    }
                    if(stack.empty()){
                        root = null;
                        break;
                    }
                }
            }
        }
        return list;
    }
}

Runtime: 384 ms

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