LeetCode 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? 递归是微不足道的，你能使用迭代完成吗？

方法一 递归

	public List<Integer> preorderTraversal(TreeNode root) {
		List<Integer> result = new ArrayList<>();
		preorderTraversal(root, result);
		return result;
	}

	public void preorderTraversal(TreeNode root, List<Integer> list) {
		if (root != null) list.add(root.val);
		else return;
		if (root.left != null) preorderTraversal(root.left, list);
		if (root.right != null) preorderTraversal(root.right, list);
	}

方法二  使用栈，非递归

	public List<Integer> preorderTraversal2(TreeNode root) {
		List<Integer> list = new ArrayList<>();
		if (root == null) return list;
		Stack<TreeNode> stack = new Stack<>();
		stack.push(root);
		while (!stack.isEmpty()) {
			TreeNode tmp = stack.pop();
			list.add(tmp.val);
			if (tmp.right != null) stack.push(tmp.right);
			if (tmp.left != null) stack.push(tmp.left);
		}
		return list;
	}