剑指offer:面试题18——树的子结构

剑指offer:面试题18——树的子结构

题目:输入两棵二叉树A,B,判断B是不是A的子结构。(ps:我们约定空树不是任意一个树的子结构)

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    bool DoesTreeHaveTree2(TreeNode *pRoot1,TreeNode*pRoot2)
    {
        if(pRoot2==NULL)   //树2到达叶子节点
            return true;
        if(pRoot1==NULL)    //树1到达叶子节点
            return false;
        if(pRoot1->val!=pRoot2->val)
            return false;
        return DoesTreeHaveTree2(pRoot1->left,pRoot2->left)&&DoesTreeHaveTree2(pRoot1->right,pRoot2->right);
    }
public:
    bool FindRoot(TreeNode *pRoot1,TreeNode*pRoot2)
    {
        bool result =false;
        
        if(pRoot2!=NULL && pRoot1!=NULL) 
        {
            if(pRoot1->val==pRoot2->val)
                result=DoesTreeHaveTree2(pRoot1,pRoot2);
            if(!result)
                result=FindRoot(pRoot1->left,pRoot2);
            if(!result)
                result=FindRoot(pRoot1->right,pRoot2);
        }
        return result;
    }
public:
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        return FindRoot(pRoot1,pRoot2);
    }
};

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转载自blog.csdn.net/MereX/article/details/89631600