572. Subtree of Another Tree

和剑指上树的子结构的题稍稍有点不同，

     3 
    / 
   4   5
  / 
 1   2    
    /
   0

   4
  / 
 1   2
这两个在这个题目中不是子树的关系，但剑指那个题这种情况算是子树

class Solution {
public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        bool result = false;
        if(s != NULL && t != NULL){
            if(s->val == t->val)
                result = isSubCore(s,t);
            if(!result)
                result = isSubtree(s->left,t);
            if(!result)
                result = isSubtree(s->right,t);
        }
        return result;
    }
    bool isSubCore(TreeNode* s,TreeNode* t){
        if(t == NULL && s == NULL)
            return true;
        else if(t == NULL && s != NULL)
            return false;
        else if(t != NULL && s == NULL)
            return false;
        if(s->val != t->val)
            return false;
        return isSubCore(s->left,t->left) && isSubCore(s->right,t->right);
    }
};