Question:
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
题意
给你一颗树的后序和中序，求层序遍历。
思路：
方法一：无需构造树，用map来取代。
首先，根据后序定根，中序定左右的关系来递归得到这棵树的先序遍历情况。
其次，只需要在先序遍历的过程中加一个变量index，表示当前的根结点在二叉树中所对应的下标（根节点从0开始，那么左子树表示为2i+1,右子树表示为2i+2），所以进行一次输出先序的递归过程中，就可以把根结点下标index及所对应的值存储在map<int, int> lev中，而又因为map是有序的，会根据index从小到大自动排序，这样递归完成后level中的值就是层序遍历的顺序。详见参考代码一。
参考代码一：
#include <bits/stdc++.h>
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define LL long long
#define x first
#define y second
#define PII pair<int,int>
#define PDD pair<double,double>
using namespace std;
const int INF=0x3f3f3f3f;
const int N=1e5+5;
const int M=1e7;
const int mod=1e9+7;
int n,first;
int pos[33],in[33];
map<int,int> lev;
void dfs(int pr,int il,int ir,int index)
{
if(il>ir)return;
lev[index]=pos[pr];
int u=il;
while(u<=ir && pos[pr]!=in[u])u++;
//左树
dfs(pr-1-(ir-u),il,u-1,index*2+1);
//右树
dfs(pr-1,u+1,ir,index*2+2);
}
int main()
{
// io;
cin>>n;
first=1;
for(int i=0;i<n;i++)
{
cin>>pos[i];
}
for(int i=0;i<n;i++)
{
cin>>in[i];
}
dfs(n-1,0,n-1,0);
auto it=lev.begin();
cout<<it->y;
while(++it!=lev.end())
{
cout<<" "<<it->y;
}
system("pause");
return 0;
}
/*
*/
