1020 Tree Traversals（25 分）
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
解题思路：将中序，后序转化成前序，并多加了一个参数，index，左子树就2*index+1，右就2*index+2，先将其赋值为-1，嘴笨，还是代码比较清楚。
#include<bits/stdc++.h>
using namespace std;
vector<int>pos,in,level(100000,-1);
void pre(int root,int start,int end,int index)
{
if(start>end) return ;
int i=start;
while(i<end&&in[i]!=pos[root]) i++;
level[index]=pos[root];
pre(root-(end-i+1),start,i-1,2*index+1);
pre(root-1,i+1,end,2*index+2);
}
int main(void)
{
int n,flag=0;
scanf("%d",&n);
pos.resize(n);
in.resize(n);
for(int i=0;i<n;i++) scanf("%d",&pos[i]);
for(int i=0;i<n;i++) scanf("%d",&in[i]);
int i=0;
pre(n-1,0,n-1,0);
for(int i=0;i<level.size();i++)
{
if(level[i]!=-1)
{
if(flag!=0) printf(" ");
printf("%d",level[i]);
}
flag++;
}
return 0;
}