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;}