1086. Tree Traversals Again (25)-蒲公英云

1086. Tree Traversals Again (25)

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.

Figure 1

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
Pop

Sample Output:

3 4 2 6 5 1

题目大意：

代码：

#include<stdio.h>
#include<stack>
using namespace std;
struct node
{
    int data;
    int lchild,rchild;
}tree[50];
int pre[100],in[100],l1=0,l2=0;
int biuld(int num,int *a,int *b)
{
    if(num<=0)
    {
        return -1;
    }
    tree[*a].data=*a;
    int i;
    for(i=0;i<num;i++)
    {
        if(*(b+i)==*a)
            break;
    }
    tree[*a].lchild=biuld(i,a+1,b);
    tree[*a].rchild=biuld(num-i-1,a+i+1,b+i+1);
    return *a;
}
int flag=0;
void posttraverl(int root)
{
    if(root!=-1)
    {
        posttraverl(tree[root].lchild);
        posttraverl(tree[root].rchild);
        if(flag==0)
        {
            printf("%d",root);
            flag=1;
        }
        else
        {
            printf(" %d",root);
        }
    }
}
int main()
{
    int i,j,n,m,k,t;
    char operation[5];
    scanf("%d",&n);
    stack<int> s;
    for(i=0;i<n*2;i++)
    {
        scanf("%s",operation);
        if(operation[1]=='u')
        {
            scanf("%d",&m);
            s.push(m);
            pre[l1++]=m;
        }
        else
        {
           in[l2++]=s.top();
           s.pop();
        }
    }
    int root=biuld(n,pre,in);
    posttraverl(root);
    return 0;
}