1127. ZigZagging on a Tree (30)

柔光的暖阳◎ 2022-05-28 11:59 225阅读 0赞

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

![nc_ol5xekjcdy4.jpg][]

**Input Specification:**

Each input file contains one test case. For each case, the first line gives a positive integer N (<= 30), the total number of nodes in the binary tree. The second line gives the inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

**Output Specification:**

For each test case, print the zigzagging sequence of the tree in a line. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

**Sample Input:**

8
    12 11 20 17 1 15 8 5
    12 20 17 11 15 8 5 1

**Sample Output:**

1 11 5 8 17 12 20 15

--------------------

题目大意：

代码：

#include<stdio.h>
    #include<stdlib.h>
    #include<queue>
    #include<stack>
    #include<vector>
    using namespace std;
    vector<int> v[50];
    int l=0;
    struct node
    {
        int data;
        struct node *lchild,*rchild;
    };
    struct node *create(int n,int a[],int b[])
    {
        if(n<=0)
        {
            return NULL;
        }
        struct node *root=(struct node *)malloc(sizeof(struct node));
        root->data=b[n-1];
        int *p;
        for(p=a;*p!=b[n-1];p++);
        int t=p-a;
        root->lchild=create(t,a,b);
        root->rchild=create(n-t-1,p+1,b+t);
        return root;
    }
    void level(struct node *root)
    {
        queue<struct node *> q;
        q.push(root);
        int lastnode=root->data;
        while(!q.empty())
        {
           struct node *tmp=q.front();
           q.pop();
           if(tmp!=NULL)
           {
               v[l].push_back(tmp->data);
               q.push(tmp->lchild);
               q.push(tmp->rchild);
               if(lastnode==tmp->data)
               {
                   if(tmp->rchild!=NULL)
                    lastnode=tmp->rchild->data;
                   else if(tmp->lchild!=NULL)
                    lastnode=tmp->lchild->data;
                   l++;
               }
           }
        }
    
    }
    void preorder(struct node *root)
    {
        if(root!=NULL)
        {
            printf("%d ",root->data);
            preorder(root->lchild);
            preorder(root->rchild);
        }
    }
    int main()
    {
        int i,j,n,m,k,t;
        int a[50],b[50];
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
        }
        struct node *root=create(n,a,b);
        level(root);
        int flag=0;
        for(i=0;i<=l;i++)
        {
            if(i%2==0)
            {
                for(j=v[i].size()-1;j>=0;j--)
                {
                    if(flag==0)
                    {
                        printf("%d",v[i][j]);
                        flag=1;
                    }
                    else
                    {
                        printf(" %d",v[i][j]);
                    }
                }
            }
            else
            {
                for(j=0;j<v[i].size();j++)
                {
                    if(flag==0)
                    {
                      printf("%d",v[i][j]);
                    }
                    else
                    {
                        printf(" %d",v[i][j]);
                    }
                }
            }
        }
        return 0;
    }

[nc_ol5xekjcdy4.jpg]: http://nos.patest.cn/nc_ol5xekjcdy4.jpg