1020. Tree Traversals (25)已知后序中序 层次遍历

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:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

#include<cstdio>#include<cstring>#include <queue>using namespace std;struct node{    int data;    struct node*l,*r;};struct node *creat(int *last, int * in, int n){   struct node *root;   for(int i = 0; i < n; i++)   {       if(in[i] == last[ n -1])       {          root = new node();          root->data = in[i];          root->l = creat(last, in, i );          root->r = creat(last+i, in+i+1, n - i -1);          return root;       }   }   return NULL;}void bfs(node *root){   queue<node*>q;   q.push(root);   int flag = 0;   while(!q.empty())   {       node * u = q.front();       q.pop();       if(flag)          printf(" ");       flag = 1;       printf("%d", u->data);       if(u->l)       {           q.push(u->l);       }       if(u->r)       {          q.push(u->r);       }   }}int main(){   int n;   scanf("%d", &n);   int last[110];   int in[110];   for(int i = 0; i < n; i++)   {       scanf("%d", &last[i]);   }     for(int i = 0; i < n; i++)   {       scanf("%d", &in[i]);   }   struct node *root = new node();   root = creat(last, in, n);   bfs(root);}