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:

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<cstdio>
#include<iostream>
#include<queue>
using namespace std;
typedef struct NODE{
    struct NODE* lchild, *rchild;
    int data;
}node;
int in[31], post[31];
int N;
node* create(int inL, int inR, int postL, int postR){
    if(postL > postR){
        return NULL;
    }
    int head = postR;
    node * root = new node;
    root->data = post[head];
    int i;
    for(i = inL; i <= inR; i++){
        if(in[i] == post[head])
            break;
    }
    int numL = i - inL;
    root->lchild = create(inL, i - 1, postL, postL + numL - 1);
    root->rchild = create(i + 1, inR, postL + numL, postR - 1);
    return root;
}
void levelOrder(node* tree){
    queue<node*> Q;
    Q.push(tree);
    node* temp;
    int cnt = 0;
    while(Q.empty() == false){
        temp = Q.front();
        cnt++;
        Q.pop();
        printf("%d", temp->data);
        if(cnt != N)
            printf(" ");
        if(temp->lchild != NULL)
            Q.push(temp->lchild);
        if(temp->rchild != NULL)
            Q.push(temp->rchild);
    }
}
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* tree = create(0, N - 1, 0, N - 1);
    levelOrder(tree);
    cin >> N;
    return 0;
}

总结：

1、题意：给出后序遍历和中序遍历，求层序遍历。

2、层序遍历时不要忘记在访问完元素后pop。

3、创建二叉树结束的条件是后序遍历的区间为空（postL > postR）。

4、注意控制最后一个空格不要输出，可以设置一个计数器，为N时不输出空格。