PAT [A1020]-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 1:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output 1:
4 1 6 3 5 7 2

解题思路：
递归构造二叉树（后序中序确定一颗二叉树），并利用BFS输出层次遍历。

AC代码：

#include <cstdio>#include <queue>using namespace std;const int maxn = 50;int N, postOrder[maxn], inOrder[maxn];struct node{    int v;    node* lChild;    node* rChild;};node* create(int lp, int rp, int li, int ri){    if (lp > rp) return NULL;    node* root = new node;    root->v = postOrder[rp];    int index;    for (int i = li; i <= ri; i++){        if (inOrder[i] == postOrder[rp]){            index = i;            break;        }    }    root->lChild = create(lp, index - 1 - li + lp, li, index - 1);    root->rChild = create(rp-ri+index, rp-1, index + 1, ri);    return root;}void BFS(node* root){    int cnt = 0, ans[maxn];    queue<node*> que;    que.push(root);    while (!que.empty()){        node *t = que.front();        ans[cnt++] = t->v;        que.pop();        if(t->lChild != NULL) que.push(t->lChild);        if (t->rChild != NULL) que.push(t->rChild);    }    for (int i = 0; i < cnt; i++){        printf("%d", ans[i]);        if (i < cnt - 1) printf(" ");    }}int main(){    freopen("C:\\Users\\Administrator\\Desktop\\test.txt", "r", stdin);    while (scanf("%d", &N) != EOF){        for (int i = 0; i < N; i++){            scanf("%d", &postOrder[i]);        }        for (int i = 0; i < N; i++){            scanf("%d", &inOrder[i]);        }        node* root = create(0, N-1, 0, N-1);        BFS(root);        printf("\n");    }    fclose(stdin);    return 0;}