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

题意：给你一棵二叉树的结构（n个结点），然后给你n个数，把树填进二叉树中，构成一棵二叉排序树。

思路：二叉排序树的中序遍历为递增序列，把二叉树结构建好后，输出中序遍历节点下标，然后再把给定的n个数递增排序。用一个数组来记录下标（0-N-1）与数的对应关系，然后再把二叉树结构层序遍历，并按下标顺序输出对应的数，既为结果。

#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
struct node
{
    int data;
    node * lchild;
    node * rchild;
};
int n;
vector<vector<int> > vec;
vector<int> in_order_num(n);
vector<int> in_num(n);
node * creat_tree(int start){
    if(start == -1){
        return NULL;
    }

    node * root = new node();
    root -> data = start;

    root -> lchild = creat_tree(vec[start][0]);
    root -> rchild = creat_tree(vec[start][1]);


    return root;



}
int q = 0;
void in_order(node * t){
    if(t == NULL){
        return;
    }
    in_order(t -> lchild);
    in_order_num[t -> data] = in_num[q];
    q++;
    in_order(t -> rchild);
}

void level_order(node *t){
    queue<node *> que;
    que.push(t);

    while (!que.empty())
    {
        printf("%s%d",que.front() == t ? "" : " ", in_order_num[que.front() -> data]);
        if(que.front() -> lchild != NULL){
            que.push(que.front() -> lchild);
        }
        if(que.front() -> rchild != NULL){
            que.push(que.front() -> rchild);
        }
        que.pop();
    }
    


}
int main(){

    cin >> n;
    vec.resize(n);
    in_order_num.resize(n);
    in_num.resize(n);
    for(int i = 0; i < n; i++){
        int l, r;
        scanf("%d %d", &l, &r);
        vec[i].push_back(l);
        vec[i].push_back(r);
    }
    for(int i = 0; i < n; i++){
  
        scanf("%d", &in_num[i]);

    }
    sort(in_num.begin(), in_num.end());
    
    node * tree = creat_tree(0);
    in_order(tree);
    level_order(tree);



    return 0;
}