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

幾種二叉樹遍歷方式轉換參考：二叉樹幾種遍歷方式之間的轉換 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>

using namespace std;
const int maxn = 1005;

int in_order[maxn],post_order[maxn];
int n,flag;
int lch[maxn],rch[maxn];
queue<int> que;

int build_tree(int l1,int r1,int l2,int r2)
{
    if(l1 > r1) return 0;
    int root = post_order[r2];                 //當前區間中的根節點
    int pos = l1;
    while(in_order[pos] != root) pos++;        //在中序遍歷中找到根節點的位置，將序列分成左右兩個區間
    int cnt = pos - l1;                        //左子樹區間的長度
    lch[root] = build_tree(l1,pos-1,l2,l2+cnt-1); //遍歷左子樹
    rch[root] = build_tree(pos+1,r1,l2+cnt,r2-1); //遍歷右子樹
    return root;
}
void BFS(int root)
{
    que.push(root);
    while(!que.empty())
    {
        int node = que.front();
        que.pop();
        if(!flag)
            printf("%d",node),flag = 1;
        else
            printf(" %d",node);
        if(lch[node]) que.push(lch[node]);
        if(rch[node]) que.push(rch[node]);
    }
}
int main()
{
    scanf("%d",&n);
    for(int i = 0;i < n;i ++)
        scanf("%d",post_order+i);
    for(int i = 0;i < n;i ++)
        scanf("%d",in_order+i);
    int root = build_tree(0,n-1,0,n-1);

    BFS(root);
    printf("\n");
}