A1127. ZigZagging on a Tree

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<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef struct NODE{
     struct NODE* lchild, *rchild;
     int data;
     int lev;
 }node;
 int post[], in[], N;
 node* create(int inL, int inR, int postL, int postR, int level){
     if(inL > inR){
         return NULL;
     }
     node* root = new node;
     root->data = post[postR];
     root->lev = level;
     int mid;
     for(mid = inL; mid <= inR; mid++){
         if(in[mid] == root->data)
             break;
     }
     int len = mid - inL;
     root->lchild = create(inL, mid - , postL, postL + len - , level + );
     root->rchild = create(mid + , inR, postL + len, postR - , level + );
     return root;
 }
 void levelOrder(node* root){
     queue<node*> Q;
     Q.push(root);
     int nowLev = root->lev;
     int reverTag = ;
     int cnt = ;
     vector<node*> vec;
     while(Q.empty() == false){
         node* temp = Q.front();
         Q.pop();
         if(temp->lev == nowLev){
             vec.push_back(temp);
         }else{
             nowLev = temp->lev;
             if(reverTag == ){
                 reverTag = ;
                 for(int i = vec.size() - ; i>= ; i--){
                     cnt++;
                     printf("%d ", vec[i]->data);
                 }
             }else{
                 reverTag = ;
                 for(int i = ; i < vec.size(); i++){
                     cnt++;
                     printf("%d ", vec[i]->data);
                 }
             }
             vec.clear();
             vec.push_back(temp);
         }
         if(temp->lchild != NULL)
             Q.push(temp->lchild);
         if(temp->rchild != NULL)
             Q.push(temp->rchild);
     }
     if(reverTag == ){
         for(int i = ; i < vec.size(); i++){
             cnt++;
             if(cnt == N)
                 printf("%d", vec[i]->data);
             else printf("%d ", vec[i]->data);
         }
     }else{
         for(int i = vec.size() - ; i >= ; i--){
             cnt++;
             if(cnt == N)
                 printf("%d", vec[i]->data);
             else printf("%d ", vec[i]->data);
         }
     }
 }
 int main(){
     scanf("%d", &N);
     for(int i = ; i <= N; i++){
         scanf("%d", &in[i]);
     }
     for(int i = ; i <= N; i++){
         scanf("%d",&post[i]);
     }
     node* root = create(, N, , N, );
     levelOrder(root);
     cin >> N;
     return ;
 }

总结：

1、题意：中序后序建树，再用层序输出，输出时一行逆序一行正序。

2、可以在建树时顺便将层次信息也加入节点。在层序遍历时使用一个vector，在当前层数相同时，仅仅把本该访问的节点存入vector，当层数发生改变时，输出vector内所有元素（设置一个计数器，如果上次是正序输出，则本次逆序输出）； 或者正常层序遍历，再多用一个栈+一个队列，当需要正序输出该层时，节点入队列，需要逆序则节点入栈，层数改变时输出栈中或队列中全部节点，并清空。