PAT甲级A1020 Tree Traversals (25)

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

题意：给定二叉树中序和后序序列，让你输出其层序遍历序列。

思路：首先根据中序和后序建立二叉树，然后层序遍历即可。

参考代码： 

#include<queue>
#include<cstdio>
using namespace std;
struct node{
	int data;
	node* lchild;
	node* rchild;
	node(int d):data(d),lchild(NULL),rchild(NULL){}
};
vector<int> in,post;
node* Create(int inl,int inr,int postl,int postr)
{
	if(postl>postr) return NULL;
	node* root=new node(post[postr]);
	int k=inl;
	while(k<=inr&&in[k]!=post[postr]) k++;
	int numleft=k-inl;
	root->lchild=Create(inl,k-1,postl,postl+numleft-1);
	root->rchild=Create(k+1,inr,postl+numleft,postr-1);
	return root;
}
void BFS(node* root)
{
	if(root==NULL) return;
	queue<node*> q;
	q.push(root);
	while(!q.empty()){
		node* p=q.front();
		q.pop();
		if(p!=root) printf(" %d",p->data);
		else printf("%d",p->data);
		if(p->lchild!=NULL) q.push(p->lchild);
		if(p->rchild!=NULL) q.push(p->rchild);
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	in.resize(n);
	post.resize(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;
}

优化：在使用中序和后序建树的过程中，每层节点都是按照从左往右的顺序依次访问，因此开设一个数组下表作为层号，将建树过程每次访问的结点依次加入，最后按层号输出即为层序遍历。

参考代码：

#include<cstdio>
using namespace std;
const int maxn=35;
vector<int> in(maxn),post(maxn);
vector<int> level[maxn];
void Create(int inl,int inr,int postl,int postr,int d)
{
	if(postl>postr) return ;
	level[d].push_back(post[postr]);
	int k=inl;
	while(k<=inr&&in[k]!=post[postr]) k++;
	int numleft=k-inl;
	Create(inl,k-1,postl,postl+numleft-1,d+1);
	Create(k+1,inr,postl+numleft,postr-1,d+1);
}
int main()
{
	int n;
	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]);
	Create(0,n-1,0,n-1,1);
	printf("%d",level[1][0]);
	for(int i=2;i<maxn;i++){
		for(int j=0;j<level[i].size();j++){
			printf(" %d",level[i][j]);
		}
	}
	printf("\n");
	return 0;
}