UVA122（BFS）

Description
Background
Trees are fundamental in many branches of computer science. Current state-of-the art parallel computers such as Thinking Machines’ CM-5 are based on fat trees. Quad- and octal-trees are fundamental to many algorithms in computer graphics.
This problem involves building and traversing binary trees.
The Problem
Given a sequence of binary trees, you are to write a program that prints a level-order traversal of each tree. In this problem each node of a binary tree contains a positive integer and all binary trees have have fewer than 256 nodes.
In a level-order traversal of a tree, the data in all nodes at a given level are printed in left-to-right order and all nodes at level k are printed before all nodes at level k+1.
For example, a level order traversal of the tree
picture28
is: 5, 4, 8, 11, 13, 4, 7, 2, 1.
In this problem a binary tree is specified by a sequence of pairs (n,s) where n is the value at the node whose path from the root is given by the string s. A path is given be a sequence of L’s and R’s where L indicates a left branch and R indicates a right branch. In the tree diagrammed above, the node containing 13 is specified by (13,RL), and the node containing 2 is specified by (2,LLR). The root node is specified by (5,) where the empty string indicates the path from the root to itself. A binary tree is considered to be completely specified if every node on all root-to-node paths in the tree is given a value exactly once.
The Input
The input is a sequence of binary trees specified as described above. Each tree in a sequence consists of several pairs (n,s) as described above separated by whitespace. The last entry in each tree is (). No whitespace appears between left and right parentheses.
All nodes contain a positive integer. Every tree in the input will consist of at least one node and no more than 256 nodes. Input is terminated by end-of-file.
The Output
For each completely specified binary tree in the input file, the level order traversal of that tree should be printed. If a tree is not completely specified, i.e., some node in the tree is NOT given a value or a node is given a value more than once, then the string “not complete” should be printed.
Sample Input
(11,LL) (7,LLL) (8,R)
(5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()
(3,L) (4,R) ()
Sample Output
5 4 8 11 13 4 7 2 1
not complete

这道题是一道BFS的题，也是我做的第一道BFS的题，搞了挺长时间的。这道题的题意是给出一系列结点的值，如果某个结点已经被复制，那就输出“not complete”，如果所有结点输入完毕后，发现从根到叶子结点之间某些结点没有被赋值，那也输出“not complete”。如果两种情况都不存在，就用BFS按层遍历，将树输出。

BFS遍历方式：

void BFS(vector<int> &ans){    ans.clear();    queue<Node*> po;    po.push(root);//先将root放入队列    while (!po.empty()) {        Node *now = po.front();        po.pop();//用完后就pop出去        ans.push_back(now->v);        if (now->lc != NULL) po.push(now->lc);//如果有左结点，就将左结点放入队列        if (now->rc != NULL) po.push(now->rc);//如果有右结点，就将右结点放入队列    }}

用队列存储下一个需要遍历的结点位置，一个结点一个结点的遍历

#include <iostream>#include <queue>#include <cstdio>#include <cstring>#include <vector>#include <iterator>using namespace std;bool failed = false;char s[3000];struct Node {    bool have_value;    Node *lc, *rc;    int v;    Node():have_value(false), lc(NULL), rc(NULL) {}};Node *root;Node *new_node(){    return new Node();}void remove_tree(Node *u){    if (u == NULL) return;    remove_tree (u->lc);    remove_tree (u->rc);    delete u;}void add_node(int vi, char *p){    int len = strlen(p);    Node *u = root;    for (int i = 0; i < len; i++) {        if (p[i] == 'R') {            if (u->rc == NULL) u->rc = new_node();            u = u->rc;        } else if (p[i] == 'L') {            if (u->lc == NULL) u->lc = new_node();            u = u->lc;        }    }    if (u->have_value) failed = true;    u->v = vi;    u->have_value = true;}bool BFS(vector<int> &ans){    ans.clear();    queue<Node*> po;    po.push(root);    while (!po.empty()) {        Node *now = po.front();        po.pop();        if (!now->have_value) return false;        ans.push_back(now->v);        if (now->lc != NULL) po.push(now->lc);        if (now->rc != NULL) po.push(now->rc);        //cout << po.empty() << endl;    }    return true;}bool read_input(){    failed = false;    remove_tree (root);    root = new_node ();//别忘了给root分配空间    while (1) {        if (scanf ("%s", s) == EOF) return false;        if (!strcmp(s, "()")) break;        int vi;        sscanf (&s[1], "%d", &vi);        add_node (vi, strchr(s, ',') + 1);    }    return true;}int main(){    #ifndef ONLINE_JUDGE    freopen ("in.txt", "r", stdin);    #endif // ONLINE_JUDGE    while (read_input()) {        vector<int> ans;        bool first = true, suc = BFS(ans);        //cout << suc << endl;        if (failed || !suc) {            printf ("not complete\n");        } else {            for (vector<int>::iterator it = ans.begin(); it != ans.end(); it++) {                printf (first ? "%d" : " %d", *it);                first = false;            }            printf ("\n");        }    }    return 0;}