题解 | #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 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);
    }
};