Source:

PAT A1020 Tree Traversals (25 分)

Description:

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 (≤), 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:

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

Sample Output:

4 1 6 3 5 7 2

Keys:

Code:

 /*
time: 2019-06-30 14:40:45
problem: PAT_A1020#Tree Traversals
AC: 08:33 题目大意:
给出后序和中序遍历,打印层序遍历
*/
#include<cstdio>
#include<queue>
using namespace std;
const int M=;
int post[M],in[M],n;
struct node
{
int data;
node *lchild,*rchild;
}; node *Create(int postL, int postR, int inL, int inR)
{
if(postL > postR)
return NULL;
node *root = new node;
root->data = post[postR];
int k;
for(k=inL; k<=inR; k++)
if(in[k] == post[postR])
break;
int numLeft = k-inL;
root->lchild = Create(postL,postL+numLeft-,inL,k-);
root->rchild = Create(postL+numLeft,postR-,k+,inR);
return root;
} void LayerOrder(node *root)
{
queue<node*> q;
q.push(root);
int pt=;
while(!q.empty())
{
root = q.front();
q.pop();
printf("%d%c", root->data, ++pt==n?'\n':' ');
if(root->lchild)
q.push(root->lchild);
if(root->rchild)
q.push(root->rchild);
}
} int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE scanf("%d", &n);
for(int i=; i<n; i++)
scanf("%d", &post[i]);
for(int i=; i<n; i++)
scanf("%d", &in[i]);
node *root = Create(,n-,,n-);
LayerOrder(root); return ;
}

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