1020. Tree Traversals

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

Input Specification:

Each input file contains the one test case. For each case, the first line gives a positive integer N (<=30) and 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 is separated by a space.

Output Specification:

for each test case, print on one line the level order traversal sequence of the CORRESP Onding binary tree. All the numbers in a line must is separated by exactly one space, and there must is no extra space at the end of the line.

#include <stdio.h> #include <stdlib.h> #include <queue>using namespace std;typedef struct node{int        Id        struct Node *left,*right; }node;    Node * creat (int * post,int ps,int pe,int * in,int is,int IE); void BFS (Node * root); int N;int main () {int i;    scanf ("%d", &n);    int *post=new Int[n];    int *in=new Int[n];         for (i=0;i<n;i++) {scanf ("%d", post+i);         } for (i=0;i<n;i++) {scanf ("%d", in+i);    } Node *root=creat (post,0,n-1,in,0,n-1);    BFS (root);    printf ("\ n");    System ("pause");    return 0;     }void BFS (Node * root) {Queue<node *>q;     Q.push (root);     Node *temp=new node;     int count=0;            while (!q.empty ()) {Temp=q.front ();            Q.pop ();//Team printf ("%d", temp->id);            count++;            if (count<n) printf ("");            if (temp->left) Q.push (temp->left);         if (temp->right) Q.push (temp->right);    }}node * creat (int * post,int ps,int pe,int * in,int is,int IE) {if (PS&GT;PE) return NULL;     Node * root=new node;     root->id=post[pe];     int i;     for (i=is;i<=ie;i++) if (in[i]==root->id) break;     int num=i-is;     Root->left=creat (post,ps,ps+num-1,in,is,i-1);     Root->right=creat (Post,ps+num,pe-1,in,i+1,ie);     return root; }

1020. Tree Traversals