1020 Tree Traversals

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:

1
2
3

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

1	4 1 6 3 5 7 2

Code

#include 
#include 
using namespace std;
int n;
struct TNode{
    int data;
    TNode *lchild;
    TNode *rchild;
};

TNode* create(int post[],int in[],int postL,int postR,int inL,int inR){
    if(postL>postR)
        return NULL;
    TNode *p = new TNode;
    p->data = post[postR];
    int pos;
    for(pos=inL;in[pos]!=post[postR]&&pos<=inr;pos++);< span>
    int num = pos-inL;
    p -> lchild = create(post,in,postL,postL+num-1,inL,pos-1);
    p -> rchild = create(post,in,postL+num,postR-1,pos+1,inR);
    return p;
}
void level_order(TNode *T){
    queue  q;
    TNode *p;
    q.push(T);
    int num=0;
    while(!q.empty()){
        p = q.front();
        printf("%d",p->data);
        num++;
        if(num
            printf(" ");
        q.pop();
        if(p->lchild != NULL) q.push(p->lchild);
        if(p->rchild != NULL) q.push(p->rchild);
    }
}
int main(){
    TNode* T;
    scanf("%d", &n);
    int post[n],in[n];
    for(int i=0;i
        scanf("%d", &post[i]);
    for(int i=0;i
        scanf("%d", &in[i]);
    T = create(post,in,0,n-1,0,n-1);
    level_order(T);
    return 0;    
}