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?

public class Solution {
    public List preorderTraversal(TreeNode root) {
       List res = new ArrayList<>();
       Stack stack = new Stack<>();
       if(root == null) return res;
       stack.push(root);
       while(!stack.isEmpty()){
           TreeNode node = stack.pop();
           res.add(node.val);
           if(node.right!=null) stack.push(node.right);
           if(node.left!=null) stack.push(node.left);
       }
       return res;
    }
}

public class Solution {
    public List preorderTraversal(TreeNode root) {
       List res = new ArrayList<>();
       Stack stack = new Stack<>();
       while(root!=null || !stack.isEmpty()){
           if(root == null){
                root = stack.pop();
           }
           else{
               res.add(root.val);
               stack.push(root.right);
               root = root.left;
           }
       }
       return res;
       
    }
}

二刷
stack

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List preorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        Stack stack = new Stack<>();
        while(!stack.isEmpty() || root != null){
            if(root == null) root = stack.pop().right;
            else{
                res.add(root.val);
                stack.push(root);
                root = root.left;
            }
        }
        return res;
    }
}