<p>Given the roots of two binary trees <code>root</code> and <code>subRoot</code>, return <code>true</code> if there is a subtree of <code>root</code> with the same structure and node values of<code> subRoot</code> and <code>false</code> otherwise.</p>

<p>A subtree of a binary tree <code>tree</code> is a tree that consists of a node in <code>tree</code> and all of this node&#39;s descendants. The tree <code>tree</code> could also be considered as a subtree of itself.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/subtree1-tree.jpg" style="width: 532px; height: 400px;" />
<pre>
<strong>Input:</strong> root = [3,4,5,1,2], subRoot = [4,1,2]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/subtree2-tree.jpg" style="width: 502px; height: 458px;" />
<pre>
<strong>Input:</strong> root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
<strong>Output:</strong> false
</pre>

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

<ul>
	<li>The number of nodes in the <code>root</code> tree is in the range <code>[1, 2000]</code>.</li>
	<li>The number of nodes in the <code>subRoot</code> tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>-10<sup>4</sup> &lt;= root.val &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= subRoot.val &lt;= 10<sup>4</sup></code></li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void serializer(TreeNode *root, std::string& res) {
        if (!root) {
            res += "#";
            return;
        }
        res += "^";
        res += std::to_string(root->val);
        serializer(root->left, res);
        serializer(root->right, res);
    }

    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        std::string one("");
        serializer(root, one);
        std::string two("");
        serializer(subRoot, two);
        // using C++ string find
        if (one.find(two) != std::string::npos) {
            return true;
        }
        return false;
       // return kmp(one, two);
    }

    // At the time of working on this 02/11,
    // I was not sure whether it is worth memorizing KMP
    bool kmp(std::string& one, std::string& two) {
        // create lps - longest prefix suffix
        std::vector<int> lps(two.size(), 0);

        for (int i = 1, j = 0; i < two.size(); i++) {
            if (two[i] == two[j]) {
                lps[i] = ++j;
            }
            else if (j > 0){
                j = lps[j-1];
            }
        }
        std::cout << "herllo";
        return true;
    }
};



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