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\n");

            printf("%d\n",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\n");

            continue;

        }



        printf("YES\n");

        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("\n");

    }

    

    return 0;

}

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