572. Subtree of Another Tree Code class Solution { bool check(TreeNode* root, TreeNode* subRoot) { if (root == nullptr && subRoot == nullptr) return true; if (subRoot == nullptr || root == nullptr) return false; if (root->val != subRoot->val) return false; return check(root->left, subRoot->left) && check(root->right, subRoot->right); } public: bool isSubtree(TreeNode* root, TreeNode* subRoot) { if (check(root, subRoot)) return true; if (root == nullptr) return false; return isSubtree(root->left, subRoot) || isSubtree(root->right, subRoot); } };