判断一棵二叉树是否是二叉树的子树

#include <iostream>

using namespace std;

struct TreeNode {
int val;
struct TreeNode *left, *rigth;
TreeNode(int x) :
val(x), left(NULL), rigth(NULL) {
}
};
class Solution {
public:
bool DoHasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
{
bool result = false;
if (pRoot2 == NULL)//能遍历到pRoot2说明至少pRoot2一支已经匹配成功
{
return true;
}
if (pRoot1 == NULL)//一定要在上面之后，如果为真，说明pRoot1所在的一支没有，匹配不成功
{
return false;
}
if (pRoot1->val != pRoot2->val)
{
return false;
}
result = DoHasSubtree(pRoot1->left, pRoot2->left);//左支匹配成功
if (result != false)//在左支匹配成功的前提下，进行右支匹配
{
result = DoHasSubtree(pRoot1->right, pRoot2->right);
}
return result;
}

bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
{
bool result = false;
if (pRoot1 != NULL && pRoot2 != NULL) //pRoot1 == NULL到达叶子节点还没找到，则返回false；pRoot2 == NULL一开始就为空直接返回false
{
if (pRoot1->val == pRoot2 ->val)//找到根节点相同，可能匹配
{
result = DoHasSubtree(pRoot1, pRoot2);//匹配
}
if (result == false)//上述匹配不成功，继续遍历pRoot1的所有左子树，反之结束遍历
{
result = HasSubtree(pRoot1->left, pRoot2);
}
if (result == false)//pRoot1的所有左子树没有匹配成功，那就看pRoot1的所有右子树
{
result = HasSubtree(pRoot1->right, pRoot2);
}
}
return result;
}
};

int main(void)
{
TreeNode p1(8), p2(8), p3(7), p4(9), p5(2), p6(4), p7(7);
TreeNode n1(8), n2(9), n3(2);
p1.left = &p2;
p1.rigth = &p3;
p2.left = &p4;
p2.rigth = &p5;
p5.left = &p6;
p5.rigth = &p7;
n1.left = &n2;
n1.rigth = &n3;
TreeNode *Head = NULL;
Solution s;
s.HasSubtree(&p1, &n1);
return 0;
}