题解 | #树的子结构#
树的子结构
http://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88
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) {
if(pRoot1==NULL||pRoot2==NULL)return false;
//通过递归去比较每一个结点，找到一个点就开始递归
return check(pRoot1,pRoot2)||HasSubtree(pRoot1->left,pRoot2)||HasSubtree(pRoot1->right,pRoot2);
}
//这里是判断是否有子结点
bool check(TreeNode* pRoot1, TreeNode* pRoot2){
//遍历完了。说明是A的子结构，已经在叶结点下面
if(pRoot2==NULL)return true;
//不匹配直接返回
if(pRoot1==NULL||pRoot1->val!=pRoot2->val)return false;
return check(pRoot1->left,pRoot2->left)&&check(pRoot1->right,pRoot2->right);
}
};
