PAT——1020 Tree Traversals (25 分)（後續+先序構建二叉樹）

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

#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;

int n;
int in[31], post[31], level[31];

typedef struct node
{
    int data;
    node *l_child;
    node *r_child;
}node;

void input()
{
    //
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        cin >> post[i];
    }
    for(int i = 0; i < n; i++)
        cin >> in[i];
}

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(int i = inL; i <= inR; i++)
    {
        if(in[i] == root->data)
        {
            k = i;
            break;
        }
    }
    int left_num = k - inL;
    root->l_child = create(postL, postL+left_num-1, inL, k-1);
    root->r_child = create(postL+left_num, postR-1, k+1, inR);
    return root;
}
void level_order(node *root)
{
    queue<node*> que;
    que.push(root);
    int len = 0;
    while(!que.empty())
    {
        node *father = que.front();
        que.pop();
        level[len++] = father->data;
        if(father->l_child != NULL)
        {
            que.push(father->l_child);
        }
        if(father->r_child != NULL)
        {
            que.push(father->r_child);
        }
    }
}
int main()
{
    /*1020 Tree Traversals (25 分)*/
    #ifndef ONLINE_JUDGE
        freopen("test.txt", "r", stdin);
    #endif // ONLINE_JUDGE

    input();
    node *root = create(0, n-1, 0, n-1);

    level_order(root);

    cout << level[0];
    for(int i = 1; i < n; i++)
        cout <<" " << level[i];
    return 0;
}