1086. Tree Traversals Again (25)

﹏ヽ暗。殇╰゛Y 2022-05-29 21:11 86阅读 0赞

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.

![bs_n9mde9jcnyj.jpg][]  
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;
    }

[bs_n9mde9jcnyj.jpg]: http://nos.patest.cn/bs_n9mde9jcnyj.jpg