PAT A1127 ZigZagging on a Tree （30 分）——二叉树，建树，层序遍历

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.

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

 

#include <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=31;
int pos[maxn],in[maxn];
int n,m;
vector<int> res[31];
struct node{
    int data;
    node* left,*right;
    int lvl;
};
node* create(int inl,int inr,int posl,int posr){
    if(inl > inr) return NULL;
    node* root = new node;
    root->data = pos[posr];
    int k;
    for(k=inl;k<=inr;k++){
        if(pos[posr]==in[k]) break;
    }
    int numleft=k-inl;
    root->left = create(inl,k-1,posl,posl+numleft-1);
    root->right = create(k+1,inr,posl+numleft,posr-1);
    return root;
}
void pr(node* root){
    queue<node*> q;
    root->lvl=0;
    q.push(root);
    while(!q.empty()){
        node* now=q.front();
        q.pop();
        res[now->lvl].push_back(now->data);
        if(now->left!=NULL){
            now->left->lvl=now->lvl+1;
            q.push(now->left);
        }
        if(now->right!=NULL){
            now->right->lvl=now->lvl+1;
            q.push(now->right);
        }
    }
}
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",&pos[i]);    
    }
    node* root = create(0,n-1,0,n-1);
    pr(root);
    int cnt=0;
    for(int i=0;i<32;i++){
        if(i%2==0){
            for(int j=res[i].size()-1;j>=0;j--){
                printf("%d",res[i][j]);
                cnt++;
                if(cnt<n)printf(" ");
            }
        }
        else{
            for(int j=0;j<res[i].size();j++){
                printf("%d",res[i][j]);
                cnt++;
                if(cnt<n)printf(" ");
            }
        }
        if(cnt==n)break;
    }
}

注意点：最简单的一道30分题了。只要会建树，再层序遍历，把每层的数据保存起来，再根据层数正着输出或是反向输出就ac了。

ps:也可以用deque双向队列更方便的实现。每层的输出其实可以不用开vector数组，一层遍历完后输出就好了。