1020. 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
提交代码

 

#include <string>
#include<algorithm>
#include <iostream>
#include<queue>
using namespace std;
int postorder[35];
int inorder[35];
int m;
typedef struct tn
{
	struct tn* letf;
	struct tn* right;
	int value;
}treenode;
queue<treenode *> qu;
treenode * createtree(int pl,int pr,int il,int ir)
{
	int p;
	treenode * tt=new treenode;
	if(ir<il||pr<pl)return NULL;

	else
	{
		tt->value=postorder[pr];
//		printf("%d ",tt->value);
		for(p=il;p<m;p++)
			if(inorder[p]==postorder[pr]) break;
//		printf("*%d*",p);
		tt->letf=createtree(pl,pl+p-il-1,il,p-1);
		tt->right=createtree(pl+p-il,pr-1,p+1,ir);
	}
	return tt;
}
	
void levelorder(treenode* t)
{
	treenode* ttt;
	int first=1;
	if(t!=NULL) qu.push(t);
	while(!qu.empty())
	{
		ttt=qu.front ();
			qu.pop();
		if(first)
			first=0;
		else printf(" ");
		printf("%d",ttt->value);
		if(ttt->letf!=NULL)
			qu.push (ttt->letf);
		if(ttt->right!=NULL)
			qu.push (ttt->right);
	}
}

int main()
{
	freopen("data.in","r",stdin);
	int n,i,j,k;
	scanf("%d",&m);

	for(i=0;i<m;i++)
	{
		scanf("%d",&postorder[i]);

	}
	for(i=0;i<m;i++)
	scanf("%d",&inorder[i]);
	treenode* root;
	root=createtree(0,m-1,0,m-1);
	levelorder(root);
	return 0;
}

　　

posted @ 2012-10-06 15:25  scjyldq  阅读(149)  评论(0)    收藏  举报