A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format left_index right_index

Output Specification:

For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42

Sample Output:

58 25 82 11 38 67 45 73 42

題意：有n個點，接下來輸入n行（0-n-1）表示左孩子有孩子節點，如果-1表示沒有，最後一行輸入9個數字，按照上面這棵樹的樣子將這9個數按照二叉搜尋樹填進去，最後輸出填進去後的層序遍歷的結果。

解題思路：二叉搜尋樹的中序遍歷就是結點值從小到大排序的結果，根據這個性質解決這題。我們用結構體儲存左右孩子以及結點值，接下來就是中序遍歷，中序遍歷中將結點值填進去，然後是層序遍歷，用一個數組儲存層序遍歷的結點值就ok了，具體的見程式碼。

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
vector<int>in,level,key;
int index=0;
struct Avl{
	int data,l,r;
}v[110];
void inOrder(int root) 
{
	
	if(v[root].l!=-1) inOrder(v[root].l);
	v[root].data=key[index++];
	if(v[root].r!=-1) inOrder(v[root].r);
}
void levelOrder(int root)
{
	queue<int>q;
	q.push(root);
	while(!q.empty())
	{
		int cur=q.front();
		q.pop();
		level.push_back(v[cur].data);
		if(v[cur].l!=-1) q.push(v[cur].l);
		if(v[cur].r!=-1) q.push(v[cur].r);
	}
}

int main(void)
{
	int n,root=0,l,r;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d %d",&v[i].l,&v[i].r);
	}
	key.resize(n);
	for(int i=0;i<n;i++)
	scanf("%d",&key[i]);
	sort(key.begin(),key.end());
	inOrder(root);
	levelOrder(root);
    for(int i=0;i<n;i++)
    printf("%d%s",level[i],i==n-1?"\n":" ");
	return 0;
}