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

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提交代码

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代码
#include

#include 
#include
#include 
#include
using namespace std;
int postorder[35];
int inorder[35];
int m;
typedef struct tn
{
	struct tn* letf;
	struct tn* right;
	int value;
}treenode;
queue qu;
treenode * createtree(int pl,int pr,int il,int ir)
{
	int p;
	treenode * tt=new treenode;
	if(irvalue=postorder[pr];
//		printf("%d ",tt->value);
		for(p=il;pletf=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

　　

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转载于:https://www.cnblogs.com/scjyldq/archive/2012/10/06/2713006.html