A10120. Tree Traversals

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;
}