1127 ZigZagging on a Tree （30 分）--PAT甲级

1127 ZigZagging on a Tree （30 分）

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.

zigzag.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：利用中序和后序建二叉树，之后以z字形层序输出。

c++

#include 
#include
#include
#include 
#include
#include
#include

using namespace std;
const int maxn = 510;
bool visit[maxn];
int n;
int post[maxn];
int in[maxn];
struct node{
    int val;
    node* left;
    node* right;
};

node* create(int inL,int inR,int postL,int postR){
    if(postL>postR) return NULL;
    node* now=new node;
    int data=post[postR];
    now->val=data;
    int i;
    for(i=inL;i<=inR;i++){
        if(in[i]==data) break;
    }
    int numLeft=i-inL;
    now->left=create(inL,i-1,postL,postL+numLeft-1);
    now->right=create(i+1,inR,postL+numLeft,postR-1);
    return now;
}

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&in[i]);
    }
    for(int i=0;i<n;i++){
        scanf("%d",&post[i]);
    }
    node* root=create(0,n-1,0,n-1);
    queue<node*>q;
    q.push(root);
    int first=root->val;
    bool flag=false;
    while(!q.empty()){
        int sum=q.size();
        if(flag){
            for(int i=0;i<sum;i++){
                node* now=q.front();
                printf(" %d",now->val);
                if(now->left) q.push(now->left);
                if(now->right) q.push(now->right);
                q.pop();
            }
            flag=false;
        }
        else{
            stack<int> s;
            for(int i=0;i<sum;i++){
                node* now=q.front();
                s.push(now->val);
                if(now->left) q.push(now->left);
                if(now->right) q.push(now->right);
                q.pop();
            }
            while(!s.empty()){
                int t=s.top();
                if(t==first) printf("%d",t);
                else printf(" %d",t);
                s.pop();
            }
            flag=true;
        }
    }
}