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 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #define inf 0x3f3f3f3fusing namespace std;int s[135], l[135];int ans[135];typedef int element;typedef struct node{    element data;    node *lchild, *rchild;}node, *Tree;Tree creat(int sl, int sr, int ll, int lr){    if(sr-sl <0)        return NULL;    Tree T;    T = new node;    T->data = s[sr];    int p=ll;    while(l[p] != s[sr])        ++p;    int t = lr - p;    T->lchild = creat(sl, sr-t-1, ll, p-1);    T->rchild = creat(sr-t, sr-1, p+1, lr);    return T;}void posd(Tree T){    if(T)    {        printf("%d ", T->data);        posd(T->lchild);        posd(T->rchild);    }}void level(Tree T){    int out = 0;    int in = 0;    Tree p[150];    p[in++] = T;    int j = 0;    while(in > out)    {        if(p[out])        {            ans[j++] = p[out]->data;            p[in++] = p[out]->lchild;            p[in++] = p[out]->rchild;        }        out++;    }}int main(){    int n, m, k;    int i, j;    cin>>n;    Tree L;    for(i = 0; i < n; i++)    {        scanf("%d", &s[i]);    }    for(i = 0; i < n; i++)    {        scanf("%d", &l[i]);    }    L = creat(0,n-1,0,n-1);    posd(L);    /*level(L);    for(i = 0; i < n-1; i++)    {        printf("%d ",ans[i]);    }    printf("%d\n", ans[n-1]);*/}

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