<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>

class Solution {
public:
    void traversal(TreeNode* a, TreeNode* b, bool& same) {
        if (!same)
            return;
        if ((a == nullptr) != (b == nullptr)) {
            same = false;
            return;
        }
        if ((a == nullptr) || (b == nullptr))
            return;
        if (a->val != b->val) {
            same = false;
            return;
        }
        traversal(a->left, b->left, same);
        traversal(a->right, b->right, same);
    }

    bool isSameTree(TreeNode* p, TreeNode* q) {
        bool same = true;
        traversal(p, q, same);
        return same;
    }

    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        if (root == nullptr) return false; // Base case

        if (isSameTree(root, subRoot)) {
            return true;
        }
        
        // Check the left and right subtrees
        return isSubtree(root->left, subRoot) || isSubtree(root->right, subRoot);
    }
};

29000