1127 ZigZagging on a Tree (30 分)

2019-05-07

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.

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:

1
2
3

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

Sample Output:

1	1 11 5 8 17 12 20 15

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一棵树的后序中序遍历，要求输出z字型层序遍历（即一层左向右，下一层右向左这样）

分析

先根据后序中序构造出树。然后层序遍历，根据层数判断方向。

代码

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <deque>
#include <queue>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int n;
vector<int> post,in;
struct node{
    int v;
    node* left;
    node* right;
};
node* build(int is,int ie,int ps,int pe){
    if(ps>pe||is>ie)
        return nullptr;
    int x=post[pe];
    int mid=is;
    while(in[mid]!=x)mid++;
    node* root=new node;
    root->v=x;
    root->left=build(is,mid-1,ps,ps+mid-is-1);
    root->right=build(mid+1,ie,ps+mid-is,pe-1);
    return root;
}

int main() {
    cin>>n;
    post.resize(n);
    in.resize(n);
    for(int i=0;i<n;i++)
        cin>>in[i];
    for(int i=0;i<n;i++)
        cin>>post[i];
    node* root=build(0, n-1, 0, n-1);
    deque<node*> q;
    vector<node*> vec;
    q.push_back(root);
    int l=0,first=0;
    while(!q.empty()){
        l++;
        int size=q.size();
        if(size==0)
            break;
        vec.clear();
        for(int i=0;i<size;i++){
            node* temp=q.front();
            q.pop_front();
            if(temp->left)
                q.push_back(temp->left);
            if(temp->right)
                q.push_back(temp->right);
            vec.push_back(temp);
        }
        if(l%2==0)
            for(int i=0;i<size;i++){
                first++==0?cout<<"":cout<<" ";
                cout<<vec[i]->v;
            }
        else
            for(int i=size-1;i>=0;i--){
                first++==0?cout<<"":cout<<" ";
                cout<<vec[i]->v;
            }
    }
    return 0;
}