HDU 1710

来源：互联网发布：国服lol2017有mac版编辑：程序博客网时间：2024/05/19 05:30

分治思想。

主要思路：把先序序列的第一个数字当作一个根，在中序序列中寻找根，则中序序列中根前面的序列为根的左子树，后面的序列为右子树。

则再通过先序序列就可以找到当前根的左右节点，递归，将树还原，再后序遍历即可。

这套代码虽然一次性AC，但写的真的很累，太冗杂。引以为戒。

#include<cstdio>struct BTnode{    int data;    BTnode* lchild;    BTnode* rchild;}node[1001];void recovery(BTnode* node,int* pre,int *mid,int preL,int preR,int midL,int midR){    if(preR-preL==1&&midR-midL==1)    {        if(pre[preL]==mid[midL]) node[pre[preL]].rchild=&node[pre[preL+1]];        else node[pre[preL]].lchild=&node[pre[preL+1]];        return;    }    int next_mid,next_preR=preL;    for(int i=midL;i<=midR;i++)    {        if(mid[i]==pre[preL])        {            next_mid=i;            break;        }        for(int j=preL;j<=preR;j++)        {            if(pre[j]==mid[i]&&next_preR<j)            {                next_preR=j;                break;            }        }    }    //printf("%d %d %d\n",next_mid,midR,next_preR);    if(next_mid!=midL) node[pre[preL]].lchild=&node[pre[preL+1]];    if(next_mid!=midR) node[pre[preL]].rchild=&node[pre[next_preR+1]];    if(preL+1<next_preR&&midL<next_mid-1) recovery(node,pre,mid,preL+1,next_preR,midL,next_mid-1);    if(next_preR+1<preR&&next_mid+1<midR) recovery(node,pre,mid,next_preR+1,preR,next_mid+1,midR);}bool first;void visit(BTnode* node){    if(node!=NULL)    {        visit(node->lchild);        visit(node->rchild);        if(first==1) {printf("%d",node->data); first=0;}        else printf(" %d",node->data);    }}void init(){    for(int i=1;i<=1000;i++)    {        node[i].data=i;        node[i].lchild=NULL;        node[i].rchild=NULL;    }}int main(){    int n;    while(~scanf("%d",&n))    {        if(n==1) {int temp; scanf("%d",&temp); printf("%d\n",temp); continue;}        init();        int pre[n],mid[n];        for(int i=0;i<n;i++)            scanf("%d",&pre[i]);        for(int i=0;i<n;i++)            scanf("%d",&mid[i]);        recovery(node,pre,mid,0,n-1,0,n-1);        first=1;        visit(&node[pre[0]]);        printf("\n");//        for(int i=1;i<=n;i++)//        {//            printf("%d\n",node[i].data);//            BTnode* temp1=node[i].lchild;//            BTnode* temp2=node[i].rchild;//            if(temp1!=NULL) printf("%d\n",temp1->data);//            else printf("0\n");//            if(temp2!=NULL) printf("%d\n",temp2->data);//            else printf("0\n");//            printf("\n");//        }    }    return 0;}

7.22重写

#include<cstdio>#include<cstring>struct BT{    int l;    int r;    int val;    BT(int a=0,int b=0,int c=0):l(a),r(b),val(c) {}}tree[2000];int Size=0;int n,pre[2000],in[2000];int findtree(int *P,int n1,int *I,int n2){    int root=Size++;    tree[root].val=P[0];    tree[root].l=-1;    tree[root].r=-1;    if(n1==1&&n2==1)        return root;    int mid;    for(mid=0;mid<n2;mid++)    {        if(I[mid]==P[0]) break;    }    if(mid!=0) tree[root].l=findtree(P+1,mid,I,mid);    if(n1-mid-1!=0) tree[root].r=findtree(P+1+mid,n1-mid-1,I+1+mid,n2-mid-1);    return root;}bool first=1;void postorder(int root){    if(tree[root].l!=-1) postorder(tree[root].l);    if(tree[root].r!=-1) postorder(tree[root].r);    if(first) {printf("%d",tree[root].val); first=0;}    else printf(" %d",tree[root].val);}int main(){    while(~scanf("%d",&n))    {        Size=0;        first=1;        for(int i=0;i<n;i++)        {            scanf("%d",&pre[i]);        }        for(int i=0;i<n;i++)        {            scanf("%d",&in[i]);        }        int root=findtree(pre,n,in,n);        postorder(root);        printf("\n");    }    return 0;}

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