1127. ZigZagging on a Tree (30)
1127. ZigZagging on a Tree (30)
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.
![](http://nos.patest.cn/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 1Sample Output:
1 11 5 8 17 12 20 15
分析:中后序确认二叉树没什么好说的,如果有问题点击这里。
题目对层序输出做了一个变形,要求左右来回扫荡输出,开始的时候想着判断当前是第几行然后根据奇偶性从左往右输入or从右往左输入
但是做的时候发现,新输入的会对原来的判定造成影响,然后左左右右的什么很麻烦
后来想到,其实只需要按原来的层序遍历,加入两个计数器判断行数,然后按奇偶左右遍历就行了
代码:
#include<iostream> #include<deque> using namespace std; struct node{ int data; struct node *lchild,*rchild; node(){ data=0; lchild=NULL; rchild=NULL; } }; int pos[40]={0}; int mid[40]={0}; int index=0,n; struct node * creat(int start,int end){ index--; struct node *p = new struct node; int i; for(i=0;i<n;i++) if(mid[i]==pos[index]) break; p->data=pos[index]; if(i+1<end) p->rchild=creat(i+1,end); if(start<i) p->lchild=creat(start,i); return p; } void level_out(struct node root){ deque<struct node> q; q.push_back(root); int times=0,cnt=1,now_cnt=0; cout<<root.data; while(q.size()){ if(q.front().lchild) { q.push_back(*q.front().lchild); now_cnt++;} if(q.front().rchild) { q.push_back(*q.front().rchild); now_cnt++;} q.pop_front(); cnt--; if(cnt==0){ times++; if(times%2) for(int i=0;i<q.size();i++) cout<<" "<<q.at(i).data; else for(int i=q.size()-1;i>=0;i--) cout<<" "<<q.at(i).data; cnt=now_cnt; now_cnt=0; } } } int main() { cin>>n; for(int i=0;i<n;i++) cin>>mid[i]; for(int i=0;i<n;i++) cin>>pos[i]; index=n; struct node root; root=*creat(0,n); level_out(root); return 0; }
结尾:这应该是我第一道,一看就有思路,并且自己独立实现的30分题吧 _(:з」∠)_ 纪念一下