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:
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
题意:中序和后序建树,然后按zigzagging order输出。
分析:层序遍历的时候将节点输出到容器中,最后输出的时候根据奇数还是偶数来输出结点
/** * Copyright(c) * All rights reserved. * Author : Mered1th * Date : 2019-02-28-14.24.50 * Description : A1127 */ #include<cstdio> #include<cstring> #include<iostream> #include<cmath> #include<algorithm> #include<string> #include<unordered_set> #include<map> #include<vector> #include<set> #include<queue> using namespace std; ; int n,post[maxn],in[maxn]; vector<int> ans[maxn]; struct Node{ int val,layer; Node* lchild; Node* rchild; Node(int _val,int _layer){ val=_val; lchild=NULL; rchild=NULL; layer=_layer; } }; ; Node* create(int inL,int inR,int postL,int postR,int layer){ if(inL>inR) return NULL; if(layer>maxlayer) maxlayer=layer; int rootVal=post[postR]; Node* root=new Node(rootVal,layer); int k; for(int i=inL;i<=inR;i++){ if(post[postR]==in[i]){ k=i; break; } } int numLeft=k-inL; root->lchild=create(inL,k-,postL,postL+numLeft-,layer+); root->rchild=create(k+,inR,postL+numLeft,postR-,layer+); return root; } void BFS(Node* root){ queue<Node*> q; q.push(root); while(!q.empty()){ Node* now=q.front(); q.pop(); ans[now->layer].push_back(now->val); if(now->lchild!=NULL) q.push(now->lchild); if(now->rchild!=NULL) q.push(now->rchild); } } int main(){ #ifdef ONLINE_JUDGE #else freopen("1.txt", "r", stdin); #endif scanf("%d",&n); ;i<n;i++){ scanf("%d",&in[i]); } ;i<n;i++){ scanf("%d",&post[i]); } Node* root=create(,n-,,n-,); BFS(root); ;i<=maxlayer;i++){ ){ printf(]); continue; } ==){ ;j<ans[i].size();j++){ printf(" %d",ans[i][j]); } } else{ ;j>=;j--){ printf(" %d",ans[i][j]); } } } ; }
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