1020. Tree Traversals (25)已知后序中序 层次遍历

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.

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

4 1 6 3 5 7 2

#include<cstdio>
#include<cstring>
#include <queue>
using namespace std;
struct node
{int data;struct node*l,*r;
};struct node *creat(int *last, int * in, int n)
{struct node *root;for(int i = 0; i < n; i++){if(in[i] == last[ n -1]){root = new node();root->data = in[i];root->l = creat(last, in, i );root->r = creat(last+i, in+i+1, n - i -1);return root;}}return NULL;
}
void bfs(node *root)
{queue<node*>q;q.push(root);int flag = 0;while(!q.empty()){node * u = q.front();q.pop();if(flag)printf(" ");flag = 1;printf("%d", u->data);if(u->l){q.push(u->l);}if(u->r){q.push(u->r);}}}
int main()
{int n;scanf("%d", &n);int last[110];int in[110];for(int i = 0; i < n; i++){scanf("%d", &last[i]);}for(int i = 0; i < n; i++){scanf("%d", &in[i]);}struct node *root = new node();root = creat(last, in, n);bfs(root);}