/**
     * 统一一下
     * @param root
     * @return
     */
    //前序
    public static List<Integer> preOrder(TreeNode root){
         List<Integer> list = new ArrayList();
         Stack<TreeNode> stack = new Stack();
         TreeNode cur = root;
         while(cur!=null || !stack.isEmpty()){
             //一直往左压入栈
             while(cur!=null){
                 list.add(cur.val);
                 stack.push(cur);
                 cur = cur.left;
             }
             cur = stack.pop();
             cur = cur.right;
         }
         return list;
    }

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


    //后序遍历，非递归
    public static List<Integer> postOrder(TreeNode root){
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> list = new ArrayList<>();
        TreeNode cur = root;
        TreeNode p = null;//用来记录上一节点
        while(!stack.isEmpty() || cur != null){
            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.peek();
//            后序遍历的过程中在遍历完左子树跟右子树cur都会回到根结点。所以当前不管是从左子树还是右子树回到根结点都不应该再操作了，应该退回上层。
//            如果是从右边再返回根结点，应该回到上层。
            //主要就是判断出来的是不是右子树，是的话就可以把根节点=加入到list了
            if(cur.right == null || cur.right == p){
                list.add(cur.val);
                stack.pop();
                p = cur;
                cur = null;
            }else{
                cur = cur.right;
            }

        }
        return list;
}