1020. Tree Traversals (序列建树)

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

7

2 3 1 5 7 6 4

1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

见之前一篇日志《序列建树（递归） 》

#include <iostream>

using namespace std;

struct LNode

{

  LNode *lchild,*rchild;

  int data;

};

int in[];

int pos[];

LNode* fun(int pos[],int f1,int r1,int in[],int f2,int r2)//生成树

{

      if(f1>r1)  return NULL;

      LNode *p=(LNode*)malloc(sizeof(LNode));

      p->lchild=NULL;

      p->rchild=NULL;

      p->data=pos[r1];

      int i;

      for(i=f2;i<=r2;i++)

            if(in[i]==pos[r1])  break;

      p->lchild=fun(pos,f1,f1+i-f2-,in,f2,i-);

      p->rchild=fun(pos,f1+i-f2,r1-,in,i+,r2);

     return p;     

}

int main()

{

      int n;

      while(cin>>n)

      {

            int i;

          for(i=;i<n;i++)

                  cin>>pos[i];

            for(i=;i<n;i++)

                  cin>>in[i];

            LNode *q=(LNode*)malloc(sizeof(LNode));

        q=fun(pos,,n-,in,,n-);

            LNode* que[];  //层次遍历

            int front=;int rear=;

            rear++;

            que[rear]=q;

        bool fir=true;

            while(front!=rear)

            {

               front++;

               if(fir)

               {cout<<que[front]->data;fir=false;}

               else cout<<" "<<que[front]->data;

               if(que[front]->lchild!=NULL)

                 que[++rear]=que[front]->lchild;

               if(que[front]->rchild!=NULL)

                 que[++rear]=que[front]->rchild; 

            }

            cout<<endl;

      }

   return ;

}