L2-006. tree traversal

In 2016, the Group Program was designed for the Tiandao competition-the simulation competition was given the central and post-order traversal to find the pre-order traversal. To solve this type of problem, you can first construct a binary tree and then... It's easy.

Given the post-and Middle-order traversal of a binary tree, please output ITS sequence traversal. Here we assume that the key values are all positive integers that are not equal to each other.

Input Format:

The first line of the input is a positive integer N (<= 30), which is the number of nodes in the binary tree. The second row shows the post-order traversal sequence. The third row shows the sequential traversal sequence. Numbers are separated by spaces.

Output Format:

Output the sequence traversal of the tree in a row. Numbers are separated by one space. No extra space is allowed at the beginning and end of a line.

Input example:

72 3 1 5 7 6 41 2 3 4 5 6 7

Output example:

4 1 6 3 5 7 2

The last node in the Post-order traversal is the root of the tree. Then the left subtree and the right subtree are recursive from the root node. Construct a binary tree. Then BFS goes over and finds the hierarchical sequence.

/* ***********************************************Author        :guanjunCreated Time  :2016/5/15 19:30:57File Name     :1.cpp************************************************ */#include <iostream>#include <cstring>#include <cstdlib>#include <stdio.h>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <iomanip>#include <list>#include <deque>#include <stack>#define ull unsigned long long#define ll long long#define mod 90001#define INF 0x3f3f3f3f#define maxn 10010#define cle(a) memset(a,0,sizeof(a))const ull inf = 1LL << 61;const double eps=1e-5;using namespace std;struct node{    int l,r;}nod[maxn];int in[maxn];int po[maxn];int build(int l1,int r1,int l2,int r2){    if(l1>r1)return 0;    int root=po[r2];    int p=l1;    while(in[p]!=root)p++;    int cnt=p-l1;    nod[root].l=build(l1,p-1,l2,l2+cnt-1);    nod[root].r=build(p+1,r1,l2+cnt,r2-1);    return root;}void bfs(int x){    queue<int>q;    q.push(x);    vector<int>v;    while(!q.empty()){        int w=q.front();q.pop();        if(w==0)break;        v.push_back(w);        if(nod[w].l!=0)q.push(nod[w].l);        if(nod[w].r!=0)q.push(nod[w].r);    }    int m=v.size();    for(int i=0;i<m;i++){        printf("%d%c",v[i],i==m-1?10:‘ ‘);    }}int main(){    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    //freopen("out.txt","w",stdout);    int n;    while(cin>>n){        for(int i=1;i<=n;i++)cin>>po[i];        for(int i=1;i<=n;i++)cin>>in[i];        build(1,n,1,n);        int root=po[n];        bfs(root);    }    return 0;}

Purple Book examples ..

L2-006. tree traversal