pat 1127. ZigZagging on a Tree

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#PAT Advanced Level #二叉树遍历

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本文提供了一道PAT-A级算法题目“ZigZagging on a Tree”的解答方案，该题要求根据给定的中序和后序遍历序列构建二叉树，并按之字形顺序输出节点值。

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原题链接：https://www.patest.cn/contests/pat-a-practise/1127

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:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <queue>
#include <stack>

using namespace std;

struct node
{
    int num;
    node *lc,*rc;
    int step;
    node() {}
    node(int num, node *lc, node *rc,int step):num(num), lc(lc), rc(rc), step(step) {}
};

int a[35],b[35];
vector<int> vt[35];
int n;

void build(int pos,int l,int r,node *&rt)
{
    int i,tmp=-1;
    for(i=l; i<=r; i++)//注意每次在l~r区间内搜，而不是在1~n区间内搜
    {
        if(a[i]==b[pos])
        {
            tmp=i;
            break;
        }
    }
    if(tmp==-1)//必须保证pos没有超过l~r的范围，，如果没有搜到，就是超过了（也可以直接判断if(l>r)）
        return ;
    //rt=(struct node *)malloc(sizeof(struct node));
    //rt->num=b[pos];
    //rt->lc=NULL;
    //rt->rc=NULL;
    rt=new node(b[pos],NULL,NULL,0);
    if(tmp>l)
        build(pos-(r-tmp)-1,l,tmp-1,rt->lc);//注意每次的pos是后序遍历（前序遍历）的根节点位置，，是辅助中序序列找根节点位置的（找左右子树根节点位置时，都要在l~r区间的基础上找）
    if(tmp<r)
        build(pos-1,tmp+1,r,rt->rc);
}

void bfs(node *rt)//因为每一层的输出顺序都不一样，所以每一层都要单独存，然后单独输出，用vector
{
    int i,j;
    int mmax=-1;
    node *s;
    queue<node*> q;
    rt->step=0;
    q.push(rt);
    while(!q.empty())
    {
        s=q.front();
        q.pop();
        vt[s->step].push_back(s->num);
        mmax=max(mmax,s->step);
        if(s->lc!=NULL)
        {
            s->lc->step=s->step+1;
            q.push(s->lc);
        }
        if(s->rc!=NULL)
        {
            s->rc->step=s->step+1;
            q.push(s->rc);
        }
    }
    printf("%d",vt[0][0]);
    for(i=1; i<=mmax; i++)
    {
        if(i&1)
        {
            for(j=0; j<(int)vt[i].size(); j++)
                printf(" %d",vt[i][j]);
        }
        else
        {
            for(j=vt[i].size()-1; j>=0; j--)
                printf(" %d",vt[i][j]);
        }
    }
    printf("\n");
    return ;
}

int main()
{
    int i;
    node *T;
    scanf("%d",&n);
    for(i=1; i<=n; i++)
        scanf("%d",&a[i]);
    for(i=1; i<=n; i++)
        scanf("%d",&b[i]);
    build(n,1,n,T);
    bfs(T);
    return 0;
}