PAT 1020
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
题意：给出后序和中序，求出层次遍历
#include <cstdio> #include <cstring> #include <queue> #include <iostream> using namespace std; const int maxn=50; struct node //节点 { int date; node* ld; node* rd; } ; int pre[maxn],in[maxn],post[maxn]; int n; //节点个数 node* create(int postl,int postr,int inl,int inr) //由后序和中序构建一颗树 { if(postl>postr) return NULL; node* root=new node; //建立一个新的节点，存放当前二叉树的根节点 root->date =post[postr]; //新节点的数据域为当前根节点的值 int k; for(k=inl;k<=inr;k++) { if(in[k]==post[postr]) break; } int numleft=k-inl; //左子树的节点个数 root->ld =create(postl,postl+numleft-1,inl,k-1); root->rd =create(postl+numleft,postr-1,k+1,inr); return root; } int num=0; //已经输出的节点个数，便于空格输出 void bfs(node* root) { queue<node*> q; q.push(root); while(!q.empty() ) { node* now=q.front() ; q.pop() ; printf("%d",now->date ); num++; if(num<n) printf(" "); if(now->ld !=NULL) q.push(now->ld ); if(now->rd !=NULL) q.push(now->rd ); } } int main() { scanf("%d",&n); for(int i=0;i<n;i++) //后序输入 scanf("%d",&post[i]); for(int i=0;i<n;i++) //中序输入 scanf("%d",&in[i]); node* root=create(0,n-1,0,n-1); bfs(root); //层序输出 return 0; }