PAT_1020. Tree Traversals

// 1020_Tree Traversals.cpp : 定義控制檯應用程序的入口點。
//1020. Tree Traversals (25)
//
//時間限制
//400 ms
//內存限制
//65536 kB
//代碼長度限制
//16000 B
//判題程序
//Standard
//作者
//CHEN, Yue
//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


#include "stdafx.h"
#include "stdio.h"
#include "iostream"
#include "string.h"
#include "stdlib.h"
#include "string"
#include "algorithm"
#include "queue"
using namespace std;

typedef struct node{

    int value;
    struct node *lchild;
    struct node *rchild;

}node;

int post[32],in[32];

node *create(int pl,int pr,int il,int ir){
    if(pl>pr) 
        return NULL;
    int t = il;
    while(in[t]!=post[pr])
        t++;

    node *binode= (node *)malloc(sizeof(node));
    binode->value = post[pr];
    binode->lchild = create(pl,pl+t-il-1,il,t-1);
    binode->rchild = create(pl+t-il,pr-1,t+1,ir);
    return binode;
}

void bfs(node *root){
    queue<node *>q;          //隊列元素用結構體指針
    node *temp =NULL;
    q.push(root);
    bool first = true;
    while(!q.empty()){
        temp = q.front();
        q.pop();
        if(temp){
            if(first){
                cout<<temp->value;
                first = false;
            }
            else{
                cout<<" "<<temp->value;
            }
            q.push(temp->lchild);
            q.push(temp->rchild);
        }
    }
    cout<<endl;
}

int main()
{
    int n;
    while(cin>>n){
        for(int i =0;i<n;i++)
            cin>>post[i];
        for(int i =0;i<n;i++)
            cin>>in[i];
        node *root;
        root = create(0,n-1,0,n-1);
        bfs(root);
    }
    return 0;
}