ZCMU-甲级-1120

1020. Tree Traversals (25)

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

【解析】

这道题的意思是给你中序和后序序列，叫你输出层次遍历，层次遍历的意思就是一层一层的输出。和求前序类似，这里就是加了个下标意思就是在该层的下标，该层下标和下同一左子树的关系就是2*i+1。

#include<iostream>
#include <cstdio>
#include<cstring>
#include<vector>
using namespace std;
vector<int> a,b,level(100010, 0);
void pre(int root, int start, int end, int index) {
if(start > end) retur
4000
n ;
int i = start;
while(i < end && b[i]!=a[root])//寻找根结点
i++;
level[index]=a[root];
pre(root - 1 - end + i, start, i - 1, 2 * index + 1);//下标表示的是层次遍历
pre(root - 1, i + 1, end, 2 * index + 2);
}
int main()
{
int n,i;
int count1=0;
scanf("%d", &n);
a.resize(n);
b.resize(n);
for(i = 0; i < n; i++)
scanf("%d", &a[i]);
for(i = 0; i < n; i++)
scanf("%d", &b[i]);
pre(n-1, 0, n-1, 0);
for(int i = 0; i < level.size(); i++) {
if(level[i] != 0&& count1!= n - 1)
{
printf("%d ", level[i]);
count1++;
}
else if(level[i]!= 0)
{
printf("%d",level[i]);
break;
}
}
return 0;
}