1020. Tree Traversals (25)

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

题意很简单，根据后序遍历和中序构造出层序遍历。

#include<iostream>
#include<fstream>
#include<vector>
#include<queue>
using namespace std;
typedef struct NODE* PtrN;
struct NODE
{
	int root;
	PtrN left;
	PtrN right;
	//NODE(): left(NULL), right(NULL) {};
};
int* postorder;
int* inorder;
int Findroot(int begin, int end, int data)
{
	int  i;
	for (i = begin; i <= end; i++)
		if (inorder[i] == data)break;
	return i;
}
int Findmid(int postbegin, int postend, int begin, int end)
{
	int i,j,flag = 0;
	for ( i = postbegin; i <= postend; i++)
	{
		flag = 0;
		for (j = begin; j <= end; j++)
		{
			if (postorder[i] == inorder[j])
			{
				flag = 1;
				break;
			}

		}
	if (flag == 0)break;
	}
	return i - 1;
}
PtrN BuildTree(int postbegin, int postend, int inbegin, int inend)
{
	if (postbegin > postend || inbegin > inend)return NULL;
	PtrN R=(PtrN)malloc(sizeof(struct NODE));
	R->root = postorder[postend];
	int NLpostbegin, NLpostend, NLinbegin, NLinend,
		NRpostbegin, NRpostend, NRinbegin, NRinend;
    int root = Findroot(inbegin, inend, postorder[postend]);
	int mid = Findmid(postbegin, postend,inbegin, root-1);
	NLpostbegin = postbegin; NLpostend = mid;
	NLinbegin = inbegin; NLinend = root - 1;
	NRpostbegin = mid + 1; NRpostend = postend-1;
	NRinbegin = root + 1; NRinend = inend;
	R->left = BuildTree(NLpostbegin, NLpostend, NLinbegin, NLinend);
	R->right = BuildTree(NRpostbegin, NRpostend, NRinbegin, NRinend);
	return R;
}
int main()
{
	
	int N,i;
	cin >> N;
	 postorder = new int[N];
	 inorder = new int[N];
	for (i = 0; i < N; i++)
		cin >> postorder[i];
	for (i = 0; i < N; i++)
		cin >> inorder[i];
	PtrN ROOT;
	ROOT = BuildTree(0,  N-1, 0, N-1);
	queue<PtrN>que;
	PtrN head;
	que.push(ROOT);
	while (!que.empty())
	{
		head = que.front();
		que.pop();
		if (head == ROOT)
			cout << head->root;
		else
			cout << " "<<head->root ;
		if (head->left != NULL)
			que.push(head->left);
		if (head->right != NULL)
			que.push(head->right);
	}
	
	return 0;


}