PAT L2-006. 树的遍历

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题目链接：https://www.patest.cn/contests/gplt/L2-006

题意：由二叉树的后序和中序求层次遍历。

二叉树的还原：用先序或后序的顺序递归中序来建树。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define FOR(i,k,n) for(int i=k;i<n;i++)#define FORR(i,k,n) for(int i=k;i<=n;i++)#define scan(a) scanf("%d",&a)#define scann(a,b) scanf("%d%d",&a,&b)#define scannn(a,b,c) scanf("%d%d%d",&a,&b,&c)#define mst(a,n)  memset(a,n,sizeof(a))#define ll long long#define N 105#define mod 1000000007#define INF 0x3f3f3f3fconst double eps=1e-8;const double pi=acos(-1.0);struct node{    int l,r;    node()    {        l=r=0;    }}tree[N];queue<int> q;int postorder[N],inorder[N];int post;int n;void build(int root,int l,int r){    int i=r;    for(;i>=l;i--)    {        if(inorder[i]==postorder[post]) break;    }    if(i<r)    {        tree[root].r=postorder[--post];        build(tree[root].r,i+1,r);    }    if(i>l)    {        tree[root].l=postorder[--post];        build(tree[root].l,l,i-1);    }}void Bfs(int root){    while(!q.empty()) q.pop();    int cnt=0;    cnt++;    printf("%d%c",root,cnt==n?'\n':' ');    q.push(root);    while(!q.empty())    {        int cur=q.front();        q.pop();        if(tree[cur].l) cnt++,printf("%d%c",tree[cur].l,cnt==n?'\n':' '),q.push(tree[cur].l);        if(tree[cur].r) cnt++,printf("%d%c",tree[cur].r,cnt==n?'\n':' '),q.push(tree[cur].r);    }}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    while(~scan(n))    {        mst(tree,0);        post=n-1;        FOR(i,0,n) scan(postorder[i]);        FOR(i,0,n) scan(inorder[i]);        int root=postorder[n-1];        build(root,0,n-1);        Bfs(root);    }    return 0;}

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