// 前序
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }

        Deque<TreeNode> stack = new LinkedList<TreeNode>();
        TreeNode node = root;
        while (!stack.isEmpty() || node != null) {
            while (node != null) {
                res.add(node.val);
                stack.push(node);
                node = node.left;
            }
            node = stack.pop();
            node = node.right;
        }
        return res;
    }
}

// 中序
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        Deque<TreeNode> stk = new LinkedList<TreeNode>();
        while (root != null || !stk.isEmpty()) {
            while (root != null) {
                stk.push(root);
                root = root.left;
            }
            root = stk.pop();
            res.add(root.val);
            root = root.right;
        }
        return res;
    }
}

// 后序
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }

        Deque<TreeNode> stack = new LinkedList<TreeNode>();
        TreeNode prev = null;
        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            /*
            这一步判断当前节点是否是最左侧的叶子节点或者第二次访问到该节点，
            如果满足这两个条件的其中一个，就可以把这个节点的值加到res中
            */
            if (root.right == null || root.right == prev) {
                res.add(root.val);
                prev = root;
                root = null;
            } else {
                stack.push(root);
                root = root.right;
            }
        }
        return res;
    }
}