144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

难度：medium

题目：给定二叉树，返回其前序遍历结点值。（不要使用递归）

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