题目

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
结尾无空行

Sample Output:

58 25 82 11 38 67 45 73 42
结尾无空行

思路

二叉排序树，给定结构建树，利用性质（二叉树的中序遍历结果就是升序序列）填充树，然后进行层次遍历即可。

根据题目输入形式，给定下标编号，静态方式实现树比较简单。

代码

#include <bits/stdc++.h>
using namespace std;

const int maxn = 103;
int len;
int indexs = 0;
int num[maxn];
vector<int> res;

struct node{
    int data;
    int leftchild;
    int rightchild;
}Node[maxn];

void Inorder(int root) {
    if (root == -1) return;
    Inorder(Node[root].leftchild);
    Node[root].data = num[indexs++];
    Inorder(Node[root].rightchild);
}

void level(int root) {
    queue<int> myqueue;
    myqueue.push(root);
    while (!myqueue.empty()) {
        int now = myqueue.front();
        res.push_back(Node[now].data);
        myqueue.pop();
        if (Node[now].leftchild != -1) myqueue.push(Node[now].leftchild);
        if (Node[now].rightchild != -1) myqueue.push(Node[now].rightchild);
    }
}

int main(){
    cin>>len;
    for (int i = 0; i < len; i++) {
        cin>>Node[i].leftchild>>Node[i].rightchild;
    }
    for (int i = 0; i < len; i++) {
        cin>>num[i];
    }
    sort(num, num+len);
    Inorder(0);
    level(0);
    for (int i = 0; i < len; i++) {
        cout<<res[i];
        if(i != len - 1) cout<<" ";
    }
}