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:

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

Sample Output:

4 1 6 3 5 7 2

#include<iostream>
#include<queue>
using namespace std;

int root,l,r,ll,rr;
int node[1002][2];
int a[10002], b[1002];
int n;

void dfs1(int &x, int l, int r, int ll, int rr)
{
if (l > r) x = 0;
for (int i = l; i <= r; i++)
{
if (a[i] == b[rr])
{
x=i;
dfs1(node[x][0], l, i - 1, ll, ll+i-l-1);
dfs1(node[x][1], i + 1, r, ll +i-l, rr - 1);
}
}
}

void dfs(int&x, int l, int r, int ll, int rr)
{
if (l > r) x = 0;
for (int i = l; i <= r; i++)
{
if (a[i] == b[rr])
{
x = i;
dfs(node[x][0], l, i - 1, ll, ll + i - l - 1);
dfs(node[x][1], i + 1, r, ll + i - l, rr - 1);
}
}
}

int main()
{

cin >> n;

for (int i = 1; i <= n; i++)
{
cin >> b[i];
}
for (int i = 1; i <= n; i++) cin >> a[i];
dfs1(root, 1, n, 1, n);
queue<int> p;
p.push(root);
while (!p.empty())
{
int q = p.front(); p.pop();
if (root != q) printf(" ");
printf("%d", a[q]);
if (node[q][0]) p.push(node[q][0]);
if (node[q][1]) p.push(node[q][1]);
}
return 0;

}