PAT 1127. ZigZagging on a Tree (30) 建树+双端队列 随后再看

1127. ZigZagging on a Tree (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. 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:

812 11 20 17 1 15 8 512 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

建过100次的树没啥可说的。感觉自己的输出有点麻烦，应该有更简单的方法。随后在看看

#include <iostream>#include<stdio.h>#include<queue>#include<algorithm>#include<vector>#include<string>#include<string.h>#include<set>#include<deque>#include<queue>using namespace std;struct{   int lc=-1;   int rc=-1;   int value;}node[50];int p=0;int a[50],b[50];int buildtree(int h1,int t1,int h2,int t2){    if(h1>t1) return -1;    if(h1==t1)    {        node[p].lc=node[p].rc=-1;        node[p++].value=a[h1];        return p-1;    }    int pp=h1;    while(a[pp]!=b[t2]) pp++;    int x=buildtree(h1,pp-1,h2,h2+pp-h1-1);    int y=buildtree(pp+1,t1,h2+pp-h1,t2-1);    node[p].lc=x;    node[p].rc=y;    node[p++].value=b[t2];    return p-1;}void levelorder(int sta){    deque<int> Q;    Q.push_back(sta);    int flag=0;    int kongge=0;    int p1=0,p2=1;    while(!Q.empty())    {        int tmp;        if(flag==1)        {            tmp=Q.front();        Q.pop_front();        }        else        {            tmp=Q.back();            Q.pop_back();        }        if(kongge==1) cout<<' ';        kongge=1;        p1++;        cout<<node[tmp].value;        if(flag==1)        {            if(node[tmp].lc!=-1) Q.push_back(node[tmp].lc);            if(node[tmp].rc!=-1) Q.push_back(node[tmp].rc);        }        else        {            if(node[tmp].rc!=-1) Q.push_front(node[tmp].rc);            if(node[tmp].lc!=-1) Q.push_front(node[tmp].lc);        }        if(p1==p2)        {           flag=(flag+1)%2;            p2=Q.size();            p1=0;        }    }}int main(){   int n;   cin>>n;    for(int i=0;i<n;i++)     cin>>a[i];     for(int i=0;i<n;i++)     cin>>b[i];    int start=buildtree(0,n-1,0,n-1);    levelorder(start);   return 0;}