题目描述：
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
题目思路：
给定一颗搜索二叉树的0-(n-1)结点的左右孩子，0结点是根节点。给定一个序列，这个序列上的数字是这棵树的结点的权值。求出这可搜索树的层次遍历的结果。
首先我们用链表的形式存储这棵树，然后中序遍历这棵树把给定序列的升序序列的权值赋给每个结点，然后bfs层次遍历输出即可。
题目代码：
#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;struct tree{int l,r,v;
};vector<tree>v(105);
int n, l, r, a[105];
int cnt = 0;
// 中序遍历
void dfs(int root){ if(v[root].l != -1)dfs(v[root].l);
// printf("t:%d ",root);v[root].v = a[cnt++];if(v[root].r != -1)dfs(v[root].r);
}
int main(){scanf("%d",&n);for(int i = 0; i < n; i++){scanf("%d%d",&l,&r);v[i].l = l; v[i].r = r;}for(int i = 0; i < n; i++){scanf("%d",&a[i]);}sort(a,a+n);dfs(0);// 层次遍历 queue<int>q;q.push(0);while(!q.empty()){int t = q.front();if(t) printf(" ");printf("%d",v[t].v);if(v[t].l != -1)q.push(v[t].l);if(v[t].r != -1)q.push(v[t].r);q.pop();}return 0;
}