Sicily 1935 二叉树重建

数据结构的简单题，对帮助理解二叉树的遍历蛮有帮助的。

#include <iostream>#include <cstring>using namespace std;struct node{char val;node * left;node * right;};node * build(int n, char * pre, char * in){    //注意strchr返回的是指向那个字符的指针，要减去头指针 if (n <= 0) return NULL;node * ans = new node;ans->val = pre[0];int order = strchr(in, pre[0]) - in;ans->left = build(order, pre+1, in);ans->right = build(n-order-1, pre+order+1, in+order+1);//这里不能用[]，要用指针的加减法 return ans;}void BFS(node * root){if (root == NULL) return;node* q[30];//q是输出序列 q[0] = root;int n = 0, o = 1;//n代表输出节点数，o代表总结点数 while (n < o){node * temp = q[n];n++;cout << temp->val;if (temp->left != NULL) q[o++] = temp->left;if (temp->right != NULL) q[o++] = temp->right;}}void deleteTree(node * root){if (root == NULL) return;if (root->left != NULL) deleteTree(root->left);if (root->right != NULL) deleteTree(root->right);delete root;}char preorder[30], inorder[30]; int main(){int t;cin >> t;while (t --){cin >> preorder >> inorder;int len = strlen(preorder);node * root = build(len, preorder, inorder);BFS(root);cout << endl; deleteTree(root);}return 0;}