Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
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 postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2
题目给出后序遍历和中序遍历，要求得到层序遍历。右后序遍历和中序遍历可以确定该二叉树，层序遍历有点特殊，需要用到广度优先搜索。
#include <iostream> #include<queue> using namespace std; struct node { int date; node* lt, * rt; }; int ps[50], in[50],la[50]; node* create(int postl,int postr,int inl,int inr) { if (postr-postl<0) { return NULL; } node* root = new node; root->date = ps[postr]; int k; for (k = inl; k <=inr; k++) { if (in[k]==ps[postr]) break; } int ln = k-inl;//左子树个数 root->lt = create(postl,postl+ln-1,inl,k-1);//建左子树 root->rt = create(postl+ln,postr-1,k+1,inr);//建右子树 return root; } int out=0,num; void layer(node* root) {//广度优先搜索 queue<node*>q; q.push(root); int i=0; while (!q.empty()) { node* n = q.front(); q.pop(); la[i++]=n->date; if (n->lt != NULL)q.push(n->lt); if (n->rt != NULL)q.push(n->rt); } } int main() { cin >> num; for (int i = 0; i < num; ++i) { cin >> ps[i]; } for (int i = 0; i < num; ++i) { cin >> in[i]; } node* root = create(0, num - 1, 0, num - 1); layer(root); for(int i=0;i<num;i++){ if(i!=0)cout<<" "; cout<<*(la+i); } return 0; }