C++基于递归和非递归算法判定两个二叉树结构是否完全相同(结构和数据都相同)

/*两个二叉树结构是否相同(结构和数据都相同) -- 递归和非递归方法经调试可运行源码及分析如下：***/#include <stdlib.h>#include <iostream>#include <queue>using std::cout;using std::cin;using std::endl;using std::queue;/*二叉树结点定义*/typedef struct BTreeNode{  char elem;  struct BTreeNode *pleft;  struct BTreeNode *pright;}BTreeNode;/*初始化二叉树节点*/BTreeNode* btree_init(BTreeNode* &bt){  bt = NULL;  return bt;}/*先序创建二叉树*/void pre_crt_tree(BTreeNode* &bt){  char ch;  cin >> ch;  if (ch == '#')  {    bt = NULL;  }  else  {    bt = new BTreeNode;    bt->elem = ch;    pre_crt_tree(bt->pleft);    pre_crt_tree(bt->pright);  }}/*递归方式:如果两个二叉树proot都为空树，则自然相同，返回true；如果两个二叉树proot一个为空树，另一个不为空树，则不相同，返回false；如果两个二叉树的数据不相等，返回false;如果两个二叉树都不为空树，则需要分别比较左右子树后，根据比较结果共同判定，只要有一个为false，则返回false。*//*递归判断两个二叉树是否相等(结构和数据)*/bool bitree_compare(BTreeNode* proot1, BTreeNode* proot2){  if (proot1 == NULL && proot2 == NULL)//都为空    return true;  if((proot1 != NULL && proot2 == NULL) ||    (proot1 == NULL && proot2 != NULL))//一个空，一个非空    return false;  /*比较元素*/  if(proot1->elem != proot2->elem)    return false;  bool left_compare = bitree_compare(proot1->pleft, proot2->pleft);  bool right_compare = bitree_compare(proot1->pright, proot2->pright);  return (left_compare && right_compare);}/*非递归方式借助队列实现实现算法：首先将给定根节点proot1和proot2都入队第一步：当两个队列未空时，分别获取两个树的当前层次中节点总数（即当前队列中节点个数），    先比较节点个数是否相同，如果不相同，则两个树自然不同；    如果节点个数相同，需要出队进行比较(数据也要比较)。如果有一个队列未空，则退出比较。第二步：如果有一个队列未空，则清空队列并返回不同。*//*非递归方式判断两个二叉树是否相等（仅仅结构）*/bool bitree_compare_leveltraverse(BTreeNode* proot1, BTreeNode* proot2){  if (proot1 == NULL && proot2 == NULL)//都为空    return true;  queue <BTreeNode*> que1;  queue <BTreeNode*> que2;  que1.push(proot1);  que2.push(proot2);  int cur_level_nodes_num1 = 0;  int cur_level_nodes_num2 = 0;  bool flag = true;  while (!que1.empty() && !que2.empty())  {    cur_level_nodes_num1 = que1.size();    cur_level_nodes_num2 = que2.size();    //节点数目不一样时flag设为false，退出循环    if (cur_level_nodes_num1 != cur_level_nodes_num2)    {      flag = false;      break;    }    int cur_level_node_count1 = 0;    int cur_level_node_count2 = 0;    while (cur_level_node_count1 < cur_level_nodes_num1 &&        cur_level_node_count2 < cur_level_nodes_num2)    {      ++cur_level_node_count1;      ++cur_level_node_count2;      proot1 = que1.front();      que1.pop();      proot2 = que2.front();      que2.pop();      /*元素数据比较*/      if(proot1->elem != proot2->elem)      {        flag = false;        break;      }      //判断左右孩子是否相同，不同肯定结构不相同，提前break      if( proot1->pleft == NULL && proot2->pleft != NULL  ||        proot1->pleft != NULL && proot2->pleft == NULL  ||        proot1->pright == NULL && proot2->pright != NULL ||        proot1->pright != NULL && proot2->pright == NULL)      {        flag = false;        break;      }      /*下一层的节点入队*/      if(proot1->pleft)        que1.push(proot1->pleft);      if(proot1->pright)        que1.push(proot1->pright);      if(proot2->pleft)        que2.push(proot2->pleft);      if(proot2->pright)        que2.push(proot2->pright);    }    if(flag == false)      break;  }  if(flag == false)  {    while(!que1.empty())      que1.pop();    while(!que2.empty())      que2.pop();    cout << "非递归：reslut is false." << endl;    return false;  }else  {    cout << "非递归：reslut is true." << endl;    return true;  }  return true;}int main(){  BTreeNode *bt1;  BTreeNode* bt2;  bool flag;  btree_init(bt1);//初始化根节点  btree_init(bt2);//初始化根节点  cout << "creat 1th tree:" << endl;  pre_crt_tree(bt1);//创建二叉树  cout << "creat 2th tree:" << endl;  pre_crt_tree(bt2);//创建二叉树  /*递归测试*/  flag = bitree_compare(bt1, bt2);  if(flag == true)    cout<< "递归：result is true." << endl;  else    cout << "递归：result is false." << endl;  /*非递归测试*/  bitree_compare_leveltraverse(bt1, bt2);  system("pause");  return 0;}

/*测试结果如下：creat 1th tree:a b c # # # d e # # f # g # #creat 2th tree:a b c # # # d e # # f # g # #递归：result is true.非递归：reslut is true.请按任意键继续. . .------------------分界线-----------------------creat 1th tree:a b c # # # d # #creat 2th tree:a b c # # # x # #递归：result is false.非递归：reslut is false.请按任意键继续. . .本例创建的二叉树形状：a b c # # # d e # # f # g # # 如下：    a  b    dc     e  f       ga b c # # # d # # 如下：    a  b    dca b c # # # x # # 如下：    a  b    xc*/