## Problem #

Source: [PAT 1099]

### Description #

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, provided that the nodes are numbered from 0 to N−1, and 0 is always the root. If one child is missing, then −1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

### 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`

## Solution #

- 题意
- 给你树的形状和序列，让你插入序列值构成一棵二叉排序树

- 解法
- 构建好二叉树之后中序遍历并赋值即可

## Code #

```
#include <iostream>
#include <algorithm>
#include <queue>
#define max_size 101
using namespace std;
int n, val[max_size], index = 0;
struct node
{
int v, left, right;
};
node tree[max_size];
void inOrder(int root)
{
if (root != -1)
{
inOrder(tree[root].left);
tree[root].v = val[index++];
inOrder(tree[root].right);
}
}
void levelOrder()
{
queue<int> q;
q.push(0);
while (!q.empty())
{
int root = q.front();
q.pop();
cout << tree[root].v;
if (tree[root].left != -1)
q.push(tree[root].left);
if (tree[root].right != -1)
q.push(tree[root].right);
if (q.size() != 0)
cout << " ";
}
}
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
cin >> tree[i].left >> tree[i].right;
for (int i = 0; i < n; i++)
cin >> val[i];
sort(val,val+n);
inOrder(0);
levelOrder();
return 0;
}
```