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 7Sample Output:
4 1 6 3 5 7 2
题解:
典型的二叉树的遍历题目,考察了二叉树的后序遍历和层次遍历。
代码:
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; const int maxn=50; struct node{ int data; node* lch; node* rch; }; int pre[maxn],in[maxn],post[maxn]; int n; node* create(int postl,int postr,int inl,int inr){ if(postl>postr) return NULL; node* root=new node; root->data=post[postr]; int k; for(k=inl;k<=inr;k++){ if(in[k]==post[postr]) break; } int numleft=k-inl; root->lch=create(postl,postl+numleft-1,inl,k-1); root->rch=create(postl+numleft,postr-1,k+1,inr); return root; } int num=0; void bfs(node* root){ queue<node*> q; q.push(root); while(!q.empty()){ node* now=q.front(); q.pop(); printf("%d",now->data); num++; if(num<n) printf(" "); if(now->lch!=NULL) q.push(now->lch); if(now->rch!=NULL) q.push(now->rch); } } int main(){ scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&post[i]); for(int i=0;i<n;i++) scanf("%d",&in[i]); node* root=create(0,n-1,0,n-1); bfs(root); return 0; }