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An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.
Output Specification:
For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:6 Push 1 Push 2 Push 3 Pop Pop Push 4 Pop Pop Push 5 Push 6 Pop PopSample Output:
3 4 2 6 5 1
1 #include<cstdio> 2 #include<stack> 3 #include<algorithm> 4 #include<iostream> 5 #include<stack> 6 #include<set> 7 #include<map> 8 #include<vector> 9 using namespace std; 10 int per[35],in[35]; 11 stack<int> s; 12 int n; 13 void Build(int *per,int *in,int num){ 14 if(num==0){ 15 return; 16 } 17 int mid=per[0]; 18 19 //cout<<mid<<endl; 20 21 int i; 22 for(i=0;i<num;i++){ 23 if(in[i]==mid){ 24 break; 25 } 26 } 27 28 //cout<<num<<endl; 29 30 Build(per+1,in,i); 31 Build(per+1+i,in+1+i,num-1-i); 32 if(num==n){ 33 printf("%d",mid); 34 } 35 else{ 36 printf("%d ",mid); 37 } 38 } 39 int main(){ 40 //freopen("D:\\INPUT.txt","r",stdin); 41 scanf("%d",&n); 42 n*=2; 43 int i,num,pi=0,ini=0; 44 string op; 45 for(i=0;i<n;i++){ 46 cin>>op; 47 if(op=="Push"){ 48 scanf("%d",&num); 49 s.push(num); 50 per[pi++]=num; 51 } 52 else{ 53 in[ini++]=s.top(); 54 s.pop(); 55 } 56 } 57 n=n/2; 58 59 /*cout<<n<<endl; 60 for(i=0;i<n;i++){ 61 cout<<per[i]<<" "; 62 } 63 cout<<endl; 64 for(i=0;i<n;i++){ 65 cout<<in[i]<<" "; 66 } 67 cout<<endl;*/ 68 69 Build(per,in,n); 70 printf("\n"); 71 return 0; 72 }
pat1086. Tree Traversals Again (25)
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原文地址：http://www.cnblogs.com/Deribs4/p/4784479.html
