UVa 122 Trees on the level(建树，层次遍历)

题意  建树并层次遍历输出  (data,pos)  pos表示改节点位置  L代表左儿子  R代表右儿子   

建树很简单  开始在根节点  遇到L往左走遇到R往右走 节点不存在就新建   走完了就保存改节点的值  输出直接bfs就行了了  

#include<cstdio>#include<cctype>#include<cstring>using namespace std;const int maxn = 300;char s[maxn][maxn];int t, n, lans;bool comp;struct Node{    int data;    Node* left, *right;    bool have;    Node(): left(NULL), right(NULL), have(false) {};};Node *ans[maxn];bool bfs(Node* root){    int fro = 0, bac = 0;    ans[fro] = root;    while(fro <= bac)    {        if(!ans[fro]->have) return false;        if(ans[fro]->left != NULL) ans[++bac] = ans[fro]->left;        if(ans[fro]->right != NULL) ans[++bac] = ans[fro]->right;        ++fro;    }    lans = bac;    return true;}int main(){    while(~scanf("%s", s[0]))    {        n = 0;        Node root, *cur = &root;        comp = true;        do        {            t = 0;            int l = strlen(s[n]);            for(int i = 1; i < l; ++i)            {                if(isdigit(s[n][i]))                {                    t = t * 10 + s[n][i] - '0';                }                else if(s[n][i] == 'L')                {                    if(cur->left == NULL)                        cur->left = new Node;                    cur = cur->left;                }                else if(s[n][i] == 'R')                {                    if(cur->right == NULL)                        cur->right = new Node;                    cur = cur->right;                }            }            if(cur->have) comp = false;            else            {                cur->data = t;                cur->have = true;            }            cur = &root;        }        while(scanf("%s", s[++n]), s[n][1] != ')');        if(comp) comp = bfs(&root);        if(comp)        {            for(int i = 0; i < lans; ++i)                printf("%d ", ans[i]->data);            printf("%d\n", ans[lans]->data);        }        else printf("not complete\n");    }    return 0;}

 Trees on the level 

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

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 1not complete