实验4：二叉树的基本操作

c++解释： new相当于malloc()函数，其他没有区别！

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#include<iostream>
using namespace std;struct tree {int data;tree* light, *ture;
};
int jie, shen,maxx;
//创建
tree* chu() {tree* head;head = new tree;cout << "请输入数值：\n";cin >> head->data;if (head->data == 0)return nullptr;jie++;cout << "请输入该"<<head->data<<"的左结点，无则输入 0 \n";head->light = chu();cout << "请输入该" << head->data << "的右结点，无则输入 0 \n";head->ture = chu();return head;
}
//前序
void printqian(tree* head) {if (head == nullptr) return;cout << head->data << ' ';printqian(head->light);printqian(head->ture);
}
//中序
void printzong(tree* head) {if (head == nullptr) return;printzong(head->light);cout << head->data << ' ';printzong(head->ture);
}
//后序
void printhou(tree* head) {if (head == nullptr) return;printhou(head->light);printhou(head->ture);cout << head->data << ' ';
}void sheng(tree* head) {if (head == nullptr) return;shen++;maxx = max(maxx, shen);sheng(head->light);sheng(head->ture);shen--;
}int mumo() {cout << "您可以进行一下操作：\n";cout << "1：建立新二叉树\n";cout << "2：前序输出二叉树\n";cout << "3：中序输出二叉树\n";cout << "4：后序输出二叉树\n";cout << "5：查看现二叉树节点树\n";cout << "6：查看深度\n";cout << "其他：退出\n";int k; cin >> k;return k;
}int main() {tree* head=nullptr;while (true) {int k = mumo();if (k < 1 && k>6) break;switch (k) {case 1:head = chu();cout << "建立完成！\n";break;case 2:cout << "前序输出为：";printqian(head);cout << endl;break;case 3:cout << "中序输出为：";printzong(head);cout << endl;break;case 4:cout << "后序输出为：";printhou(head);cout << endl;break;case 5:cout <<"二叉树的节点有" << jie <<"个" << endl;break;case 6:maxx = -1;sheng(head);cout <<"该二叉树深度为：" << maxx << endl;break;}}return 0;
}