UVA 122 Trees on the level 二叉树的层序遍历
题目链接: UVA 122
题目描述: 输入数值和路径, 输出一棵树的层序遍历, 如果树不合法就输出-1
解题思路: 纯码力, 学会了利用sscanf来赋值, 还有就是记住可以把任意“指向字符的指针”看成是字符串, 收获很大
代码:
#include <iostream> #include <queue> #include <string> #include <vector> #include <algorithm> #include <list> #include <iterator> #include <cmath> #include <cstring> #include <forward_list> using namespace std; const int maxn = 1000+10; struct Node { int v; bool have_value; Node * left; Node * right; Node() { left = right = NULL; have_value = false; v = 0; } }node[maxn]; Node * root; bool failed; Node * newnode() { return new Node(); } void addnode(int value, char * s) { Node * u = root; int n = (int)strlen(s); for(int i = 0; i < n; i++) { if(s[i] == 'L') { if(u->left == NULL) { u->left = newnode(); } u = u->left; } else if(s[i] == 'R'){ if(u->right == NULL) { u->right = newnode(); } u = u->right; } } if(u->have_value) failed = true; u->v = value; u->have_value = true; } bool read_input() { failed = false; root = newnode(); char s[10]; while(1) { if(scanf("%s", s)!=1) return false; if(!strcmp(s,"()")) break; int v; sscanf(&s[1], "%d", &v); addnode(v, strchr(s,',')+1); } return true; } bool bfs(vector<int> & ans) { queue<Node *> q; ans.clear(); q.push(root); while(!q.empty()) { Node * u = q.front(); q.pop(); if(!u->have_value) return false; ans.push_back(u->v); if(u->left!=NULL) q.push(u->left); if(u->right!=NULL) q.push(u->right); } return true; } int main() { while(read_input()) { if(failed == true) { printf("-1\n"); continue; } vector<int> ans; if(bfs(ans)==false) { printf("-1\n"); continue; } else { for(auto i : ans) { cout << i << " "; } cout << endl; } } return 0; }
思考: 自己的代码能力还是差, 昨天的CF还没有补题吧孩子, 然后就是好好搞ACM吧, 这个东西对自己以后肯定是有好处的啊
posted on 2017-11-30 20:42 FriskyPuppy 阅读(145) 评论(0) 编辑 收藏 举报