UVA.122 Trees on the level(二叉树 BFS)
题意分析
给出节点的关系,按照层序遍历一次输出节点的值,若树不完整,则输出not complete
代码总览
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#define nmax 10000
using namespace std;
bool failed;
struct node{
bool isvalue;
node * left, *right;
int val;
node():isvalue(false),left(NULL),right(NULL){}
};
node* root;
//node* roott;
char s[nmax];
node* newnode()
{
return new node();
}
bool addnode( int v, char * str)
{
int len = strlen(str);
node*t = root;
for(int i = 0; i<len; ++i){
//if(len == 1)
if(str[i] == 'L'){
if(t->left == NULL) t->left = new node();
t = t->left;
}else if(str[i] == 'R'){
if(t->right == NULL) t->right = new node();
t = t->right;
}
}
if(t->isvalue == true) failed = true;
t->val = v;
t->isvalue = true;
return true;
}
bool read_input()
{
failed = false;
root = newnode();
for(;;){
if(scanf("%s",s) !=1) return false;
else{
int v;
if(!strcmp(s,"()")) break;
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* t = q.front();q.pop();
if(t ->isvalue == false ) {failed = true; break;}
ans.push_back(t->val);
if(t){
if(t->left)q.push(t->left);
if(t->right)q.push(t->right);
}
}
return true;
}
int main()
{
//freopen("in.txt","r",stdin);
vector<int> ans;
//roott = newnode();
while(read_input()){
bfs(ans);
if(failed) printf("%s
","not complete");
else{
bool flag = false;
for(int i = 0; i<ans.size();++i){
if(i == 0) printf("%d",ans[i]);
else printf(" %d",ans[i]);
}
printf("
");
}
}
return 0;
}