PAT-A-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:

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

Sample Output:

4 1 6 3 5 7 2

#include<iostream>#include<cstring>#include<queue>#include<algorithm>using namespace std;const int maxn = 50;struct node{  int date;  node *lchild, *rchild;};int in[maxn], post[maxn];int n;node *create(int postl, int postr, int inl, int inr){  if (postl > postr)    return NULL;  node *root = new node;  root->date = post[postr];  int k;    for (k = inl; k <= inr; k++)  {    if (in[k] == post[postr])      break;  }  int len = k - inl;  root->lchild = create(postl, postl + len - 1, inl, k - 1);  root->rchild = create(postl + len, postr - 1, k + 1, inr);  return root;}int num = 0;void layervisit(node *root){  queue<node *> q;  q.push(root);  while (!q.empty())  {    node *top = q.front();    q.pop();    num++;    cout << top->date;    if (num < n)      cout << " ";    if (top->lchild != NULL)      q.push(top->lchild);    if (top->rchild != NULL)      q.push(top->rchild);  }}int main(){  cin >> n;  for (int i = 0; i < n; i++)    cin >> post[i];  for (int i = 0; i < n; i++)    cin >> in[i];    node *root = create(0, n - 1, 0, n - 1);  layervisit(root);  system("pause");  return 0;}