PAT(Advanced Level)Practice 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 ) N\(≤100) N (≤100) which is the total number of nodes in the tree. The next N N 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 0 0 to N − 1 N-1 N−1, and 0 is always the root. If one child is missing, then −1 will represent the NULL child pointer. Finally N N 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

に言及

二叉木の構造と同じ数字を含まないシーケンスを与えると、これらの数字をこの木のノードに埋め込み、BSTにする方法は1つしかありません.この木の層序遍歴を求めます.

構想

BSTの中順遍歴は秩序化されており,この性質を利用してキーワードをノードに埋め込むことができる.

コード#コード#

#include 
#include 
#include 

using namespace std;

struct {
    int val;
    int lChild, rChild;
} node[105];

int arr[105];

int inOrder(int root, int &index) {
    if (node[root].lChild != -1)
        inOrder(node[root].lChild, index);

    node[root].val = arr[index++];

    if (node[root].rChild != -1)
        inOrder(node[root].rChild, index);
}

void levelOrder() {
    queue<int> q;
    q.push(0);

    while (not q.empty()) {
        int top = q.front();
        q.pop();

        cout << node[top].val;

        if (node[top].lChild != -1) q.push(node[top].lChild);
        if (node[top].rChild != -1) q.push(node[top].rChild);

        if (not q.empty()) cout << " ";
    }
}

int main() {
    int n;
    cin >> n;

    for (int i = 0; i < n; ++i)
        cin >> node[i].lChild >> node[i].rChild;

    for (int i = 0; i < n; ++i)
        cin >> arr[i];

    sort(arr, arr + n);

    int i = 0;
    inOrder(0, i);

    levelOrder();
}