'commit'

TangDou/LuoGuBook/P1229_Prepare.cpp

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+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 1e6 + 10;
+
+//树的结构体+存储数组
+//此处：为二叉树的标准创建、遍历模板，可用于其它试题！
+struct Node {
+    int id;     // 当前结点ID
+    int left;   // 左结点ID
+    int right;  // 右结点ID
+    char value; //当前结点的value值
+} t[N];
+
+//前序遍历
+void pre_order(Node node) {
+    //利用递归前序输出二叉树
+    if (node.id) {
+        cout << node.value << " ";
+        pre_order(t[node.left]);
+        pre_order(t[node.right]);
+    }
+}
+
+//中序遍历
+void in_order(Node node) {
+    //利用递归前序输出二叉树
+    if (node.id) {
+        in_order(t[node.left]);
+        cout << node.value << " ";
+        in_order(t[node.right]);
+    }
+}
+
+
+//后序遍历
+void post_order(Node node) {
+    //利用递归前序输出二叉树
+    if (node.id) {
+        post_order(t[node.left]);
+        post_order(t[node.right]);
+        cout << node.value << " ";
+    }
+}
+
+//层序遍历
+void level_order(Node node) {
+    queue<Node> q;
+    //放入第一个
+    q.push(node);
+    while (!q.empty()) {
+        Node n1 = q.front();
+        q.pop();
+        cout << n1.value << " ";
+        if (n1.left > 0) q.push(t[n1.left]);
+        if (n1.right > 0) q.push(t[n1.right]);
+    }
+}
+
+int n, m;
+
+/**
+  测试数据：
+  4 3
+  a 2 0
+  b 3 0
+  c 0 0
+
+  a 2 0
+  b 0 3
+  c 0 0
+
+  a 0 2
+  b 3 0
+  c 0 0
+
+  a 0 2
+  b 0 3
+  c 0 0
+
+  https://blog.csdn.net/qq_40720919/article/details/100065207
+  测试数据
+  1 7
+  A 2 5
+  B 3 4
+  D 0 0
+  E 0 0
+  C 6 7
+  F 0 0
+  G 0 0
+
+  经与网络上的例子数据对比，目前的结构体+数组的方式，确定可以解决二叉树的问题，并且不用new
+ */
+int main() {
+    cin >> n >> m;
+    //n组数据
+    for (int i = 1; i <= n; i++) {
+        //清空结构体数组,这个玩意好暴力啊
+        memset(&t, 0, sizeof(t));
+
+        //创建树
+        for (int j = 1; j <= m; j++) {
+            cin >> t[j].value >> t[j].left >> t[j].right;
+            t[j].id = j;
+        }
+        cout << endl;
+        cout << "树" << i << ":" << endl;
+        //前序
+        cout << "前序：";
+        pre_order(t[1]);
+        cout << endl;
+        //中序
+        cout << "中序：";
+        in_order(t[1]);
+        cout << endl;
+        //后序
+        cout << "后序：";
+        post_order(t[1]);
+        cout << endl;
+        //层序
+        cout << "层序：";
+        level_order(t[1]);
+        cout << endl;
+        cout << "====================================================" << endl;
+        cout << endl;
+    }
+    return 0;
+}