UVA.122 Trees on the level(二叉树 BFS)

最新推荐文章于 2020-02-23 10:04:56 发布

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#二叉树 #bfs #遍历

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本文介绍了一个基于UVA.122 Trees on the level的二叉树层序遍历问题，并提供了完整的C++实现代码。通过输入节点之间的关系，程序能够进行层序遍历并输出节点值，同时判断树是否完整。

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\n","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("\n");
        }
    }
    return 0;
}