【剑指Offer】【树】树的子结构

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:

    bool isSubtree(TreeNode *pRoot1,TreeNode* pRoot2)
    {
        if(pRoot2 == nullptr)
        {
            return true;
        }
        //不能先判断pRoot1
        if(pRoot1 == nullptr)
        {
            return false;
        }
        if(pRoot1->val == pRoot2->val)
        {
            return isSubtree(pRoot1->left,pRoot2->left) && isSubtree(pRoot1->right,pRoot2->right);
        }
        else
        {
            return false;
        }
    }
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        if(pRoot1 == nullptr || pRoot2 == nullptr)
        {
            return false;
        }
        
        //短路运算符，只要有一个正确就不会进入下一个了
        return isSubtree(pRoot1,pRoot2) || HasSubtree(pRoot1->left,pRoot2) || HasSubtree(pRoot1->right,pRoot2);
    }
    
};

//HasSubtree的另一种写法
bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        if(pRoot1 == nullptr || pRoot2 == nullptr)
        {
            return false;
        }
        
        bool ret = false;
        if(pRoot1->val == pRoot2->val)
        {
            ret = isSubtree(pRoot1,pRoot2);
        }
        if (!ret)
        {
            ret = HasSubtree(pRoot1->left, pRoot2);
        }
        if (!ret)
        {
            ret = HasSubtree(pRoot1->right, pRoot2);
        }
        return ret;
    }