http://zhedahht.blog.163.com/blog/static/25411174201011445550396/
思路：
2个递归，总体来说思路还是比较自然的。但是如何把代码写的优雅，包括一些边界的检查，还是一件需要雕琢的事情。
文章中关于子树判断的递归处理的很好，代码十分精简：
bool DoesTree1HaveAllNodesOfTree2(TreeNode* pTreeHead1, TreeNode* pTreeHead2) { if(pTreeHead2 == NULL) return true; if(pTreeHead1 == NULL) return false; if(pTreeHead1->m_nValue != pTreeHead2->m_nValue) return false; return DoesTree1HaveAllNodesOfTree2(pTreeHead1->m_pLeft, pTreeHead2->m_pLeft) && DoesTree1HaveAllNodesOfTree2(pTreeHead1->m_pRight, pTreeHead2->m_pRight); }
struct TreeNode { int m_value; TreeNode* m_lhs; TreeNode* m_rhs; }; bool try2solve(TreeNode* pHead1, TreeNode* pHead2) { if (pHead2 == NULL) return true; if (pHead1 == NULL) return false; if (pHead1->m_value != pHead2->m_value) return false; return try2solve(pHead1->m_lhs, pHead2->m_lhs) && try2solve(pHead1->m_rhs, pHead2->m_rhs); } bool HasSubTree(TreeNode* pTreeHead1, TreeNode* pTreeHead2) { if (pTreeHead1 == NULL || pTreeHead2 == NULL) return false; bool result = false; if (pTreeHead1->m_value == pTreeHead2->m_value) result = try2solve(pTreeHead1, pTreeHead2); if (!result && pTreeHead1->m_lhs) result = HasSubTree(pTreeHead1->m_lhs, pTreeHead2); if (!result && pTreeHead1->m_rhs) result = HasSubTree(pTreeHead1->m_rhs, pTreeHead2); return result; }