BST 遍历

来源：互联网发布：centos 7 乱码编辑：程序博客网时间：2024/05/19 02:44

递归遍历

pre_order

// 递归先序遍历  void rpre_order(btree* tree) {      btNode * p = tree;      if (NULL == p) {          return;      }      cout << p->data << "  ";      rpre_order(p->left);      rpre_order(p->right);  }

pos_order

// 递归后续遍历  void rpost_order(btree* tree) {      btNode* p = tree;      if (NULL == p)          return;      rpost_order(p->left);      rpost_order(p->right);      cout << p->data << "  ";  }

in_order

// 递归中序遍历  void rin_order(btree* tree) {      btNode* p = tree;      if (NULL == p)          return;      rin_order(p->left);      cout << p->data << "  ";      rin_order(p->right);  }

非递归遍历

pre

// 非递归前序遍历  void pre_order(btree* tree) {      btNode * p = tree;      if (NULL == p)          return;      stack<btNode*>* st = new stack<btNode*>();      st->push(p);      while (!st->empty()) {          p = st->top();          st->pop();          cout << p->data << "  ";          if (NULL != p->right) {              st->push(p->right);          }          if (NULL != p->left) {              st->push(p->left);          }      }      delete st;  }

pos

// 非递归后续遍历  //需要注意的是 : 非递归<strong>后序</strong>遍历方法相对实现比较困难，主要原因就在于父子节点访问缺乏连续  //              解决思路是将前一个访问节点记忆下来，并判断和当前节点之间的关系再做处理  void post_order(btree* tree) {      btNode* curr = tree;  // 记录当前执行的节点      btNode* prev = NULL; // 记录前一个访问的节点      if (NULL == curr) {          return;      }      stack<btNode*>* st = new stack<btNode*>();      st->push(curr);      while (!st->empty()) {          curr = st->top();          // 如果该节点没有孩子节点则可以直接访问          if (NULL == curr->left && NULL == curr->right) {              cout << curr->data << "  ";              st->pop();              prev = curr;              continue;          }          // 如果该节点的孩子节点已经被访问过了，也可直接访问          if (NULL != prev && (prev == curr->right || prev == curr->left)) {              cout << curr->data << "  ";              st->pop();              prev = curr;              continue;          }          // 除了上述情况则需要将孩子节点入栈          if (NULL != curr->right) {              st->push(curr->right);          }          if (NULL != curr->left) {              st->push(curr->left);          }      }      delete st;  }

in

// 非递归中序遍历  void in_order(btree* tree) {      btNode* p = tree;      if (NULL == p)          return;      stack<btNode*>* st = new stack<btNode*>();      while (NULL != p || !st->empty()) {          while (NULL != p) {              st->push(p);              p = p->left;          }          if (!st->empty()) {              p = st->top();              st->pop();              cout << p->data << "  ";              p = p->right;          }      }      delete st;  }

逐层遍历

广度优先

void bfs_order(btree* tree) {      btNode* p = tree;      if (NULL == p) {          return;      }      queue<btNode*> * qu = new queue<btNode*>();      qu->push(p);      while (!qu->empty()) {          p = qu->front();          qu->pop();          cout << p->data << "  ";          if (NULL != p->left) {              qu->push(p->left);          }          if (NULL != p->right) {              qu->push(p->right);          }      }      delete qu;  }

销毁二叉树

void destory_btree(btree* tree) {      btNode* p = tree;      if (NULL == p) {          return;      }      queue<btNode*> * qu = new queue<btNode*>();      qu->push(p);      while (!qu->empty()) {          p = qu->front();          qu->pop();          if (NULL != p->left) {              qu->push(p->left);          }          if (NULL != p->right) {              qu->push(p->right);          }          free(p);      }      delete qu;  }

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