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
Code:
#include "cstdio"
#include "vector"
#include "queue"
#include "algorithm"
using namespace std;
const int maxn = 50;
int post[maxn], in[maxn]; //后序序列和中序序列
int n; //结点个数
struct node
{
int data;
node* lchild;
node* rchild;
};
//建树,后序序列区间为[postL,postR],中序序列区间为[inL,inR]
node* create(int postL, int postR, int inL, int inR)
{
if(postL > postR)
{
return NULL;
}
node* root = new node;
root->data = post[postR]; //根结点值为后序序列最后一个值
int k;
for(k = inL; k < inR; k++)
{
if(in[k] == post[postR])
break;
}
int numLeft = k - inL;
//返回根结点左子树
root->lchild = create(postL, postL + numLeft - 1, inL, k - 1);
//返回根结点右子树
root->rchild = create(postL + numLeft, postR - 1, k + 1, inR);
return root;
}
int num = 0; //记录输出的结点数,因为输出要做判断使最后一个元素输出没有空格
//层序遍历
void BFS(node* root)
{
queue<node*> q;
q.push(root); //根结点入队
while(! q.empty()) //队不为空则出队
{
node* now = q.front(); //取队首元素
q.pop();
printf("%d",now->data);
num++;
if(num < n) printf(" "); //不是最后一个元素,则追加一个空格
if(now->lchild != NULL) q.push(now->lchild); //左子树不为空则入队
if(now->rchild != NULL) q.push(now->rchild); //右子树不为空则入队
}
}
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;
}