剑指 Offer 37. 序列化二叉树

// Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if(root == null) return "[]";
        Deque<TreeNode> deque = new LinkedList<>();
        deque.addLast(root);
        StringBuilder sb = new StringBuilder();
        sb.append("[");
        while (!deque.isEmpty()){
            TreeNode node = deque.pollFirst();
            if(node!=null) {
                sb.append(node.val);
                sb.append(",");
                deque.addLast(node.left);
                deque.addLast(node.right);
            }else{
                sb.append("null,");
            }
        }
        sb.deleteCharAt(sb.length()-1);  //删除结尾的逗号
        //将结尾的null删除
        while (sb.lastIndexOf("null")==sb.length()-4){
            sb.delete(sb.length()-5,sb.length());
        }
        sb.append("]");
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if(data.equals("[]")) return null;
        data = data.substring(1,data.length()-1);
        String[] datas = data.split(",");
        ArrayList<TreeNode> list = new ArrayList<>();
        for (String s : datas) {  //将字符串转换为整数
            if (s.equals("null")) {
                list.add(null);
            } else {
                int x = Integer.parseInt(s);
                list.add(new TreeNode(x));
            }
        }
        int len = list.size();
        TreeNode head = list.get(0);
        if(len == 1) return head;
        int i = 0,j = 1;
        while (j<len){
            if(list.get(i) == null) {
                i++;
                continue;
            }
            list.get(i).left = list.get(j);
            if(j+1<len){
                list.get(i).right = list.get(j+1);
            }
            j+=2;
            i++;
        }
        return head;
    }

这道题感觉算不上是难题