UVA 122 trees on the LEVEL--YHX

The title is as follows: Given a sequence of binary trees, you is 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 has fewer than .

In a level-order traversal of a tree, the data in all nodes at a given level is printed in left-to-right order a nd all nodes at level K is 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 was specified by a sequence of pairs (n,s) where n was the value at The node whose path from the root was given by the string s. A path is given a sequence of l' s and R's where l indicates a left branch and R Ind Icates a right branch. In the tree diagrammed above, the node containing are specified by (13,RL), and the node containing 2 are 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 being completely specified if every node on all root-to-node paths in the tree is GIV En a value exactly once.

1#include <cstdio>2#include <cstring>3#include <queue>4 using namespacestd;5 structnode6 {7     intLch,rch,val;8     BOOLb;9}a[260],n1,n2;Ten intn,ans[260]; One intRd () A { -     intI,j,k,p,x,y,z,rt; -     Chars[ -],c1,c2; theMemset (A,0,sizeof(a)); -n=1; -     if(SCANF ("%s", s) ==-1)return 0; -rt=1; +      while(1) -     { +         if(s[1]==')') Break; Ax=0; at          for(i=1; s[i]!=','; i++) -x=x*Ten+s[i]-'0'; -p=1; -          for(i=i+1; I<=strlen (s)-2; i++) -           if(s[i]=='L') -           { in               if(!a[p].lch) a[p].lch=++N; -p=A[p].lch; to           } +           Else -           { the               if(!a[p].rch) a[p].rch=++N; *p=A[p].rch; $           }Panax Notoginseng         if(a[p].b) rt=-1; -A[p].val=x; thea[p].b=1; +scanf"%s", s); A     } the     returnRT; + } -queue<int>Q; $ intMain () $ { -     inti,j,k,l,m,p,x,y,z; -     BOOLb; the      while(1) -     {Wuyix=Rd (); the         if(!x) Break; -         if(x==-1)  Wu         { -printf"Not complete\n"); About             Continue; $         } -          while(!q.empty ()) Q.pop (); -Q.push (1); -memset (ans,0,sizeof(ans)); Ak=b=0; +          while(!q.empty ()) the         { -n1=A[q.front ()]; $ Q.pop (); the             if(!n1.b) the             { theb=1; the                  Break; -             } inans[++k]=N1.val; the             if(N1.lch) Q.push (N1.lch); the             if(N1.rch) Q.push (n1.rch); About         } the         if(b) theprintf"Not complete\n"); the         Else +         { -printf"%d", ans[1]); the              for(i=2; i<=k;i++)Bayiprintf"%d", Ans[i]); theprintf"\ n"); the         } -     } -}

If the position is represented directly by an array subscript, the subscript will become large when all nodes are connected to a line. So it is necessary to record his left and right son position at each node, so as to save space.

Each node is assigned a value with a bool record, and if it is not assigned a value or is assigned a second value, it is not complete.

But the attention to detail is due to multiple sets of data, even if the read-in has been known not to complete also read.

UVA 122 trees on the LEVEL--YHX