<p>Given an integer <code>n</code>, return all the numbers in the range <code>[1, n]</code> sorted in lexicographical order.</p>

<p>You must write an algorithm that runs in&nbsp;<code>O(n)</code>&nbsp;time and uses <code>O(1)</code> extra space.&nbsp;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 13
<strong>Output:</strong> [1,10,11,12,13,2,3,4,5,6,7,8,9]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 2
<strong>Output:</strong> [1,2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class Solution {
    class Node{
        public:
        Node* child[10];
        bool isEnd;
        Node(){
            for(int i=0;i<10;i++){
                child[i] = NULL;
            }
            isEnd=false;
        }
    };
    void insertData(Node* root,string word){         // O(n)
        for(char i:word){
            if(root->child[i-'0']==NULL){
                Node* temp=new Node();
                root->child[i-'0']=temp;
            }
            root=root->child[i-'0'];
        }
        root->isEnd=true;
    }
    void help(Node* root,vector<int>& vec,string& s){
        if(root->isEnd){
            vec.push_back(stoi(s));
        }
        for(int i=0;i<10;i++){
            if(root->child[i]){
                s.push_back('0'+i);
                help(root->child[i],vec,s);
                s.pop_back();
            }
        }
    }
public:
    vector<int> lexicalOrder(int n) {
        Node* root=new Node();
        for(int i=1;i<=n;i++){

            insertData(root,to_string(i));
        }
        vector<int> res;
        string temp="";
        help(root,res,temp);
        return res;
    }
};
26401