1099 Build A Binary Search Tree （30 分）

1099 Build A Binary Search Tree （30 分）

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

思路：
　　这个题目要求输出二叉搜索树的层序遍历。我的思路是先建树，然后层次遍历即可。怎么建立一颗二叉搜索树呢？根据二叉搜索树的中序
遍历结果是有序的，所以进行中序遍历建立二叉树。题目中的输入告诉了我们二叉树的形状（每个结点的左右子女索引值都告诉我们了），所以
只需要对最后的一个key值序列进行排序，然后中序遍历赋值即可。

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<set>
using namespace std;
int tree[101];
//对树进行深度遍历
struct Node
{
    int data;
    int lchild;
    int rchild;
};
int k=0;
void inOrder(Node node[],int key[],int i)
{
    if(i!=-1)
    {
        inOrder(node,key,node[i].lchild);
        node[i].data=key[k++];
        inOrder(node,key,node[i].rchild);
    }
}
vector<int> print;
void level(Node node[])
{
    queue<int> q;
    q.push(0);
    while(!q.empty())
    {
        int temp=q.front();
        q.pop();
        print.push_back(node[temp].data);
        if(node[temp].lchild!=-1)
            q.push(node[temp].lchild);
        if(node[temp].rchild!=-1)
            q.push(node[temp].rchild);
    }
}



int main()
{
    int n;
    cin>>n;
    Node node[n];
    for(int i=0;i<n;i++)
    {
        cin>>node[i].lchild>>node[i].rchild;
    }
    int key[n];
    for(int i=0;i<n;i++)
        cin>>key[i];
    sort(key,key+n);
    inOrder(node,key,0);
    level(node);
    cout<<print[0];
    for(int i=1;i<print.size();i++)
        cout<<" "<<print[i];
    return 0;
}