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 (≤), 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<algorithm>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<cstring>
#include<queue>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn=10010;
int post[maxn];
int in[maxn];
int level[maxn];

void f(int root,int start,int end,int index) {//通过后序遍历根节点，确定中序遍历根节点下标，存入层序遍历数组中，再递归中序的左和右子树
//root 后序遍历根节点位置 //start 中序遍历序列首地址，end： 中序遍历序列尾地址   index ：满二叉树中层次结点的下标（根节点在层次序列中的位置）
    if(start>end)//中序序列长度为0，该子树遍历结束
    {
        return ;
    }
    int i=start;//i为中序遍历中根节点的下标
    while(i<end&&in[i]!=post[root]){//通过后序根节点，确定中序根节点下标
        i++;
    }
    level[index]=post[root];//根节点位置存入层序遍历数组
    f(root-(end-i+1),start,i-1,2*index+1);//递归遍历左子树
    f(root-1,i+1,end,2*index+2);//递归遍历右子树
}

int main() {
    int N;
    memset(level,-1,sizeof(level));
    scanf("%d",&N);
    for(int i=0; i<N; i++) {
        scanf("%d",&post[i]);
    }
    for(int i=0; i<N; i++) {
        scanf("%d",&in[i]);
    }
    f(N-1,0,N-1,0);
    //如果不是满二叉树，index的值非连续，所以数组的下标非连续
    int cnt=0;
    for(int i=0;i<maxn; i++) {
        if(level[i]!=-1){
            if(cnt<N-1){
                printf("%d ",level[i]);
            }
            else{
                printf("%d\n",level[i]);
            }
            if(cnt>N){
                break;
            }
            cnt++;
        }
    }
    return 0;
}