树的遍历——pat1043

http://pat.zju.edu.cn/contests/pat-a-practise/1043

给予N个数字组成二叉搜索树，判断这个数列是否由先序遍历得出或是镜像先序遍历得出，若是则输出相应的后续遍历或是镜像后续遍历

分析：

镜像先序遍历其实就是 先访问祖先，再访问右子树，再左子树

镜像后续遍历就是先访问右子树，再左子树，在祖先

#include<stdio.h>

struct tree{
    int v;
    tree *left,*right;
};

int shu[1099];
int pre[1099];
int post[1099];
int step;

void bulid(tree* head,int v){
    if(v >= head->v){
        if(head->right==NULL){
            head->right=new tree;
            head=head->right;
            head->v=v;
            head->left=NULL;
            head->right=NULL;
        }else{
            bulid(head->right,v);
        }
    }else{
        if(head->left==NULL){
            head->left=new tree;
            head=head->left;
            head->v=v;
            head->left=NULL;
            head->right=NULL;
        }else{
            bulid(head->left,v);
        }
    }
}

void preTree(tree* head,int mirror){
    pre[step]=head->v;
    step++;
    if(mirror==0){
        if(head->left!=NULL)preTree(head->left,mirror);
        if(head->right!=NULL)preTree(head->right,mirror);
    }else{
        if(head->right!=NULL)preTree(head->right,mirror);
        if(head->left!=NULL)preTree(head->left,mirror);
    }
}

void postTree(tree* head,int mirror){
    if(mirror==0){
        if(head->left!=NULL)postTree(head->left,mirror);
        if(head->right!=NULL)postTree(head->right,mirror);
    }else{
        if(head->right!=NULL)postTree(head->right,mirror);
        if(head->left!=NULL)postTree(head->left,mirror);
    }
    post[step]=head->v;
    step++;
}

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        int i,mirror=-1;
        
        tree* rhead;
        tree* head=new tree;
        rhead=head;
        
        for(i=1;i<=n;i++){
            scanf("%d",&shu[i]);
        }
        if(n==1){
            printf("YES
");
            printf("%d
",shu[1]);
            continue;
        }
        
        head->v=shu[1];
        head->left=NULL;
        head->right=NULL;
        for(i=2;i<=n;i++){
            bulid(rhead,shu[i]);
        }

        step=1;
        preTree(rhead,0);//普通的先序遍历
        for(i=1;i<=n;i++){
            if(shu[i]!=pre[i])break;
        }
        if(i==n+1)mirror=0;

        step=1;
        preTree(rhead,1);//镜像先序遍历
        for(i=1;i<=n;i++){
            if(shu[i]!=pre[i])break;
        }
        if(i==n+1)mirror=1;

        if(mirror==-1){
            printf("NO
");
            continue;
        }

        printf("YES
");
        step=1;
        postTree(rhead,mirror);
        int ok=0;
        for(i=1;i<=n;i++){
            if(ok==0)ok=1;
            else printf(" ");
            printf("%d",post[i]);
        }
        printf("
");
    }
    
    return 0;
}