UVA_122: Trees on the level

Description

Background

Trees are fundamental in many branches of computer science. Current state-of-the art parallel computers suchas Thinking Machines' CM-5 are based onfat trees. Quad- andoctal-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 printsa level-order traversal of each tree. In this problem each node of abinary tree contains a positive integer and all binary trees havehave fewer than 256 nodes.In alevel-order traversal of a tree, the data in all nodes at agiven level are printed in left-to-right order and all nodes at levelk are printed before all nodes at levelk+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) wheren is the value at the node whose path from the root isgiven by the strings. A path is given be a sequence of L's andR's where L indicates a left branch and R indicates aright branch. In the tree diagrammed above, the node containing 13 isspecified 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 thepath from the root to itself. A binary tree is considered to becompletely specified if every node on all root-to-node paths inthe 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 describedabove separated by whitespace. The last entry in each tree is (). Nowhitespace appears between left and right parentheses.All nodes contain a positive integer. Every tree in the input willconsist of at least one node and no more than 256 nodes. Input isterminated by end-of-file.

Output

For each completely specified binary tree in the input file, the levelorder traversal of that tree should be printed. If a tree is notcompletely specified, i.e., some node in the tree is NOT given a valueor a node is given a value more than once, then the string ``notcomplete'' 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

分析：

这道是一个直接的二叉树广度遍例问题，按照题目要求来处理就可以了。要注意所给的二元组可能没有数字，也可能没有路径串，这些情况都要很仔细的处理。此外，重复给出的节点以及未给出的节点都算输入错误，全部要输出not complete。在提交你的解答前请务必自己造几个这样的特殊情况检验一下，否则很可能WA。

此外，顺便提一下广度遍例的方法。需要建立一个链表（为加快插入和删除操作）来存储每一层的节点。先将根节点加入链表，然后开始依次（从左至右）遍例链表中的每一个节点。遇到一个不为空的节点，则输出之并将其删除，然后将它的两个子节点在原位置上插入，再向右继续下一个节点。一直到该层所有节点遍例结束，此时链表中的节点即为下一层的所有节点，再从头开始遍例即可。如此往复，一直到链表为空，则说明整棵树遍例结束。算法时间复杂度为O(n)。

#include <iostream>#include <cstring>#include <cstdio>#include <queue>#include <vector>using namespace std;struct Node{bool have_value;int v;Node *left, *right;Node():have_value(false),v(0),left(NULL),right(NULL) {}};const int maxn = 300;char s[maxn];bool failed;Node *root;vector<int> ans;Node* newnode(){ return new Node(); }void remove_tree(Node* u) {if(u==NULL) return ;remove_tree(u->left);remove_tree(u->right);delete u;}void addnode(int v, char* s){int n = strlen(s);Node *u = root;for(int i=0; i<n; i++){if(s[i]=='L'){if(u->left==NULL) u->left = newnode();u = u->left;}else if(s[i]=='R'){if(u->right==NULL) u->right = newnode();u = u->right;}}if(u->have_value) failed = true;u->v = v;u->have_value = true;}bool read_input(){failed = false;remove_tree(root);root = newnode();while(1){if(scanf("%s",s)!=1)return false;if(!strcmp(s,"()"))break;int v;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* u = q.front(); q.pop();if(!u->have_value) return false;ans.push_back(u->v);if(u->left != NULL) q.push(u->left);if(u->right != NULL) q.push(u->right);}return true;}int main(){while(read_input()){if(!failed && bfs(ans) && !ans.empty()){for(int i=0; i<ans.size(); i++){if(i>0)putchar(' ');printf("%d",ans[i]);}putchar('\n');}else printf("not complete\n");}    return 0;}