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[], post[];
 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 - , postL, postL + numL - );
     root->rchild = create(i + , inR, postL + numL, postR - );
     return root;
 }
 void levelOrder(node* tree){
     queue<node*> Q;
     Q.push(tree);
     node* temp;
     int cnt = ;
     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 = ; i < N; i++)
         scanf("%d", &post[i]);
     for(int i = ; i < N; i++)
         scanf("%d", &in[i]);
     node* tree = create(, N - , , N - );
     levelOrder(tree);
     cin >> N;
     return ;
 }

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。