PAT甲级1127 - ZigZagging on a Tree

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通过中序和后序进行建树,在建树的过程中,记录当前深度,同一深度的PUSH到同一个vector容器中。因为建树过程中,LR是有序的,故同一层的左右顺序已经确定了。

#include<bits/stdc++.h>
using namespace std;
const int maxn=30+2;
int inOrder[maxn],postOrder[maxn];
vector<int> ans[maxn];

void dfs(int inL,int inR,int postL,int postR,int depth){
    if(inL>inR)return;
    int i;
    for(i=inL;i<=inR;++i){
        if(inOrder[i] == postOrder[postR])break;
    }
    ans[depth].push_back(inOrder[i]);
    dfs(inL,i-1,postL,i-1-inL+postL,depth+1);
    dfs(i+1,inR,postR-inR+i,postR-1,depth+1);
}

void output(){
    int depth=0;
    bool flag=false;
    while(!ans[depth].empty()){
        if(depth%2){
            for(vector<int>::iterator it = ans[depth].begin();it!=ans[depth].end();++it){
                if(!flag)printf("%d",*it);
                else printf(" %d",*it);
                flag=true;
            }
        }else{
            for(vector<int>::reverse_iterator it = ans[depth].rbegin();it!=ans[depth].rend();++it){
                if(!flag)printf("%d",*it);
                else printf(" %d",*it);
                flag=true;
            }
        }
        ++depth;
    }
}

int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;++i)scanf("%d",inOrder+i);
    for(int i=0;i<n;++i)scanf("%d",postOrder+i);
    dfs(0,n-1,0,n-1,0);
    output();
    return 0;
}

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转载自blog.csdn.net/Mr0cheng/article/details/79415574