PAT Advanced 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

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

这题考察了给定左子树，右子树，然后再给出数字，按照给定进行构建树，最后输出层序遍历。

我们先对其进行排序，就是中序的结果。

我们仅需要在构建的时候，进行加一个level进行代表层级，用一个index代表索引，然后利用level和index进行排序即可。

最后按次序输出就是所得结果

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct node {
    int index, val, left, right, level;
};
bool cmp(node& n1, node& n2){
    return n1.level == n2.level ? n1.index < n2.index: n1.level < n2.level;
}
int N;
vector<node> v;
vector<int> data;int k = 0;
void inorder(int root, int index, int level){
    if(v[root].left != -1) inorder(v[root].left, 2 * index + 1, level + 1);
    v[root] = {index, data[k++], v[root].left, v[root].right, level};
    if(v[root].right != -1) inorder(v[root].right, 2 * index + 2, level + 1);
}
int main(){
    cin >> N;
    v.resize(N);
    data.resize(N);
    for(int i = 0; i < N; i++)
        cin >> v[i].left >> v[i].right;
    for(int i = 0; i < N; i++)
        cin >> data[i];
    sort(data.begin(), data.end());
    inorder(0, 0, 0);
    sort(v.begin(), v.end(), cmp);
    cout << v[0].val;
    for(int i = 1; i < N; i++)
        cout << " " << v[i].val;
    system("pause");
    return 0;
}