UVA 548 根据中序和后序建立二叉树并求根到叶的最短路

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题意：根据中序和后序建立二叉树并求根到叶的最短路

#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<sstream>#include<algorithm>#define inf 0x3f3f3f3fusing namespace std;const int MAXN=10010;int l[MAXN];int r[MAXN];int inorder[MAXN];int postorder[MAXN];int minx,pos;int build(int L1,int R1,int L2,int R2)//中序[L1,R1]和后序[L2,R2]建立一棵二叉树，返回根节点序号{    if(L1>R1||L2>R2)        return 0;    int root=postorder[R2];    int p=L1;    while(inorder[p]!=root)        p++;    int cnt=p-L1；;    l[root]=build(L1,p-1,L2,L2+cnt-1);    r[root]=build(p+1,R1,L2+cnt,R2-1);    return root;}void dfs(int t,int sum){    sum+=t;    if(l[t]==0&&r[t]==0)//搜到叶节点就判断    {        if(sum<minx)        {            minx=sum;            pos=t;        }        return ;    }    if(l[t]>0)        dfs(l[t],sum);    if(r[t]>0)        dfs(r[t],sum);}int main(){    string str;    while(getline(cin,str))    {        stringstream ss;        ss<<str;        int a;        int k=0;        while(ss>>a)        {            inorder[k++]=a;        }        getline(cin,str);        stringstream sx;        sx<<str;        k=0;        while(sx>>a)        {            postorder[k++]=a;        }        int root=build(0,k-1,0,k-1);        minx=inf;        dfs(root,0);        cout<<pos<<endl;    }    return 0;}

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