1020. Tree Traversals (25)

1020. Tree Traversals (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

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

题意：

1.输入一个正整数N表示二叉树有N个节点
2.输入后序的顺序
3.输入中序的顺序
4.输出层次遍历的顺序

坑：

层次的输出需从左到右方可

思路：

1.后序：左->右->根
2.中序：左->根->右
3.由此可知，根节点就是后序的最后一个节点，而根据得到的根节点，可由中序得知左右子树，递归可得树的全貌
- 例如题中所给样例：
- 后序：2 3 1 5 7 6 4
- 中序：1 2 3 4 5 6 7
- 拆分：根：4 左子树：1 2 3 右子树：5 6 7
       - 左子树拆分：根：1 左子树：空 右子树：2 3
        - 右子树拆分：根：6 左子树：5 右子树：7
4.层次顺序的存储方法是数组(要开得够大)，将根节点存储在下标为location的位置后，其左子树根节点存储在下标为location*2，右子树则是location*2+1；

代码：

#include<iostream>#include<cstring>#include<algorithm>using namespace std;int postorder[50],inorder[50];int level[10000];void findCeng(int ps,int pe,int is,int ie,int location){    if(pe<0||ie<0||ps>pe||is>ie)return;    int root=postorder[pe];    level[location]=root;    int im=0,pm=0;     if(ps==pe)return;    for(int i=is;i<=ie;i++)    {        if(inorder[i]==root)        {            im=i;            break;        }       }    for(int i=ps;i<=pe;i++)    {        bool flag=false;        for(int j=is;j<im;j++)        {            if(postorder[i]==inorder[j])            {                flag=true;                break;            }        }        if(!flag)        {            pm=i;            break;        }    }    findCeng(ps,pm-1,is,im-1,location*2);    findCeng(pm,pe-1,im+1,ie,location*2+1);}int main(){    int N;    cin>>N;    memset(level,0,sizeof(level));    for(int i=0;i<N;i++)cin>>postorder[i];    for(int i=0;i<N;i++)cin>>inorder[i];    findCeng(0,N-1,0,N-1,1);    int count=0;    for(int i=1;;i++)    {        if(level[i]!=0)        {            count++;            if(count==N)            {                cout<<level[i]<<endl;                break;            }            else cout<<level[i]<<" ";        }    }    return 0;}

附上更简洁的代码（别的大大写的哈）

    #include <cstdio>      #include <cstring>      using namespace std;      int n;      int pod[35];      int iod[35];      int lod[10000];      void dfs(int p1, int p2, int q1, int q2, int od){          if(p1 > p2 || q1 > q2) return ;          int i = p1;          while(iod[i] != pod[q2]) i++;          lod[od] = pod[q2];            dfs(p1, i-1, q1, q1+i-1-p1, 2*od);          dfs(i+1, p2, q1+i-p1, q2-1, 2*od+1);      }      int main(){          scanf("%d",&n);          memset(lod,0,sizeof(lod));          for(int i = 0; i < n; i++){              scanf("%d",&pod[i]);          }          for(int i = 0; i < n; i++){              scanf("%d",&iod[i]);          }          dfs(0, n-1, 0, n-1, 1);          int flag = 1;          for(int i = 1; i < 10000; i++){              if(flag && lod[i] != 0){                  printf("%d",lod[i]);                  flag = 0;              }               else if(!flag && lod[i]){                  printf(" %d",lod[i]);              }          }          return 0;      }