<p>Given a string <code>s</code> and a dictionary of strings <code>wordDict</code>, add spaces in <code>s</code> to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in <strong>any order</strong>.</p>

<p><strong>Note</strong> that the same word in the dictionary may be reused multiple times in the segmentation.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;catsanddog&quot;, wordDict = [&quot;cat&quot;,&quot;cats&quot;,&quot;and&quot;,&quot;sand&quot;,&quot;dog&quot;]
<strong>Output:</strong> [&quot;cats and dog&quot;,&quot;cat sand dog&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;pineapplepenapple&quot;, wordDict = [&quot;apple&quot;,&quot;pen&quot;,&quot;applepen&quot;,&quot;pine&quot;,&quot;pineapple&quot;]
<strong>Output:</strong> [&quot;pine apple pen apple&quot;,&quot;pineapple pen apple&quot;,&quot;pine applepen apple&quot;]
<strong>Explanation:</strong> Note that you are allowed to reuse a dictionary word.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;catsandog&quot;, wordDict = [&quot;cats&quot;,&quot;dog&quot;,&quot;sand&quot;,&quot;and&quot;,&quot;cat&quot;]
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 20</code></li>
	<li><code>1 &lt;= wordDict.length &lt;= 1000</code></li>
	<li><code>1 &lt;= wordDict[i].length &lt;= 10</code></li>
	<li><code>s</code> and <code>wordDict[i]</code> consist of only lowercase English letters.</li>
	<li>All the strings of <code>wordDict</code> are <strong>unique</strong>.</li>
	<li>Input is generated in a way that the length of the answer doesn&#39;t exceed&nbsp;10<sup>5</sup>.</li>
</ul>

class Solution {
public:
    vector<string> ans;
    struct Node{
        Node* list[26];
        bool isWord;
        Node(){
            for(int i=0;i<26;i++)list[i] = 0;
            isWord = false;
        }
    };

    void insert_word(Node* root, string &word){
        Node* cur = root;
        for(auto &c:word){
            if(cur->list[c-97]==0)cur->list[c-97] = new Node();
            cur = cur->list[c-97];
        }
        cur->isWord = true;
    }

    void helper(string &s, int id, string &a, Node* root){
        if(s.size() == id){ans.push_back(a.substr(1,a.size()-1));}
        Node* cur = root;
        int i;
        int k = a.size();
        a += " ";
        for(i=id;i<s.size();i++){
            if(cur->list[s[i]-97] == 0)break;
            cur = cur->list[s[i]-97];
            a.push_back(s[i]);
            if(cur->isWord){helper(s, i+1, a, root);}
        }
        a = a.substr(0,k);
    }

    vector<string> wordBreak(string s, vector<string>& wordDict) {
        Node* root = new Node();
        for(auto &word:wordDict)insert_word(root, word);
        string a = "";
        helper(s, 0, a, root);
        return(ans);
    }
};
8800