劍指offer-樹的子結構

輸入兩棵二叉樹A，B，判斷B是不是A的子結構。（ps：我們約定空樹不是任意一個樹的子結構）

python

class Solution:
    def HasSubtree(self, pRoot1, pRoot2):
        # write code here
        result=False
        if pRoot1 and pRoot2:
        	if pRoot1.val==pRoot2.val:   #先找到根節點
        		result=self.DoesTreeHaveTree(pRoot1,pRoot2)
        	if not result:
        		result=self.HasSubtree(pRoot1.left,pRoot2)
        	if not result:
        		result=self.HasSubtree(pRoot1.right,pRoot2)
        return result
        	
    def DoesTreeHaveTree(self,pRoot1,pRoot2):
    	if pRoot2==None:
    		return True
    	if pRoot1==None:
    		return False
    	if pRoot1.val != pRoot2.val:
    		return False
    	return self.DoesTreeHaveTree(pRoot1.left,pRoot2.left) and \
               self.DoesTreeHaveTree(pRoot1.right,pRoot2.right)

c++

struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        bool result = false;
	    if (pRoot1 != nullptr && pRoot2 != nullptr)
	    {
		    if (pRoot1->val==pRoot2->val)
			    result = DoesTreeHaveTree2(pRoot1, pRoot2);
		    if (!result)
			    result = HasSubtree(pRoot1->left, pRoot2);
		    if (!result)
			    result = HasSubtree(pRoot1->right, pRoot2);
	    }
	    return result;
    }
    
    bool DoesTreeHaveTree2(TreeNode* pRoot1, TreeNode* pRoot2)
    {
	    if (pRoot2 == nullptr)
		    return true;
	    if (pRoot1 == nullptr)
		    return false;
	    if (!(pRoot1->val == pRoot2->val))
		    return false;
	    return DoesTreeHaveTree2(pRoot1->left, pRoot2->left) && DoesTreeHaveTree2(pRoot1->right, pRoot2->right);
    }
};