UVA 122

Trees are fundamental in many branches of computer science (Pun definitely intended). Current stateof-
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.
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 on the right
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.
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.
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

题意不说了 贴个VJ的网址点击打开链接

直接上代码

#include <bits/stdc++.h>
using namespace std;


#define LL long long
#define INF 0x3f3f3f3
#define pi acos(-1)
const int maxn=1e3+5;
const int maxx=1e6+5;

struct tree
{
    int    V;
    tree* L;
    tree* R;
}tnode;
tree node[300];
tree*root;
int  tree_size,complete;

int test(tree *root)
{
    if (!root->V) return 0;
    int ans = 0;
    if (!root->L || test(root->L))
        ans ++;
    if (!root->R || test(root->R))
        ans ++;
    return ans;
}

void madetree(tree *root, char* str, int v)
{
    if (*str == 'R') {
        if (!root->R)
            root->R = &node[++ tree_size];
        madetree(root->R, str+1, v);
    }else if (*str == 'L') {
        if (!root->L)
            root->L = &node[++ tree_size];
        madetree(root->L, str+1, v);
    }else {
        if (root->V)
            complete = 0;
        root->V = v;
    }
}

tree*Q[300];
void output(tree *root)//相当于队列 存下所有分支
{
    int move = 0,save = 0;
    Q[save ++] = root;
    printf("%d",root->V);
    while (move < save) {
        tree* now = Q[move ++];
        if (now->L) {
            printf(" %d",now->L->V);
            Q[save ++] = now->L;
        }
        if (now->R) {
            printf(" %d",now->R->V);
            Q[save ++] = now->R;
        }
    }printf("\n");
}

int main()
{
    char buf[256],leaf[256];
    int  value;

    complete = 1;
    root = &node[tree_size = 0];//创建根结点  用newnode() 也可以
    while (~scanf("%s",buf)) {
        if (!strcmp(buf,"()")) {//判断结束标识
            if (!complete || !test(root))
                printf("not complete\n");
            else output(root);

            memset(node, 0, sizeof(node));//输出之后重置
            root = &node[tree_size = 0];
            complete = 1;
        }else {
            sscanf(buf,"(%d,%s",&value,leaf);
            madetree(root, leaf, value);
        }
    }
    return 0;
}