码迷,mamicode.com
首页 > 其他好文 > 详细

1020. Tree Traversals (序列建树)

时间:2015-01-28 14:32:56      阅读:152      评论:0      收藏:0      [点我收藏+]

标签:

 

 

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[31];

int pos[31];

 

 

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-f2-1,in,f2,i-1);

      p->rchild=fun(pos,f1+i-f2,r1-1,in,i+1,r2);

     return p;     

}

 

int main()

{

      int n;

      while(cin>>n)

      {

            int i;

          for(i=0;i<n;i++)

                  cin>>pos[i];

            for(i=0;i<n;i++)

                  cin>>in[i];

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

        q=fun(pos,0,n-1,in,0,n-1);

 

            LNode* que[40];  //层次遍历

            int front=0;int rear=0;

            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 0;

}

 

 

 

1020. Tree Traversals (序列建树)

标签:

原文地址:http://www.cnblogs.com/xiaoyesoso/p/4255582.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!