257. Binary Tree Paths (dfs recurive & stack)

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:
 
   1
 /   \
2     3
 \
  5
 
Output: ["1->2->5", "1->3"]
 
Explanation: All root-to-leaf paths are: 1->2->5, 1->3

Idea: traverse solution (inorder postorder, preorder can not solve this problem) 1253

dfs is the solution.

Basic structure:

if(node.left != null){
            sb.append("->"); sb.append(node.left.val);
            traverse(node.left, sb);
            sb.setLength(sb.length()-2 - String.valueOf(node.left.val).length()); //****
        }

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
//regular trwverse
//1 2 5 3
class Solution {
    List<String> res = new ArrayList<>();
    public List<String> binaryTreePaths(TreeNode root) {
        if(root == null) return res;
        traverse(root, (new StringBuilder()).append(root.val));
        return res;
    }
    public void traverse(TreeNode node, StringBuilder sb){
        if(node.left == null && node.right==null){
            res.add(sb.toString());//append the string
            return;
        }
        //left branch
        if(node.left != null){
            sb.append("->"); sb.append(node.left.val);
            traverse(node.left, sb);
            sb.setLength(sb.length()-2 - String.valueOf(node.left.val).length()); //****
        }
        if(node.right != null){
            sb.append("->"); sb.append(node.right.val);
            traverse(node.right, sb);
            sb.setLength(sb.length()-2 - String.valueOf(node.right.val).length());
        }
    }
}

Question: can I write it into the stack(non-recursive)?