LintCode-劍指Offer-(245)子樹

class Solution {
public:
    /**
    * @param T1, T2: The roots of binary tree.
    * @return: True if T2 is a subtree of T1, or false.
    */
   bool isSubtree(TreeNode *T1, TreeNode *T2) {
        if ( T2 == NULL )return true;
        if ( T1 == NULL )return false;
        bool result = false;
        result = TreeCmp(T1, T2);
        //將所有結果或起來就可以，因爲只要有一個地方是子樹，就爲子樹，但這樣貌似會比較慢
        return isSubtree(T1->left, T2) || isSubtree(T1->right, T2) || result;
    }
    //TreeCmp用來比較當前是否是當前節點開始的子樹，這個函數很重要
    bool TreeCmp (TreeNode *T1, TreeNode *T2) {
        if (T1 != nullptr && T2!=nullptr && T1->val == T2->val) {
            return TreeCmp(T1->left,T2->left) && TreeCmp (T1->right,T2->right);
        }
        if (T1 == nullptr && T2 == nullptr) {
            return true;
        }
        return false;
    }
};