• 【PAT甲级】1099 Build A Binary Search Tree


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    🔑中文翻译:构建二叉搜索树
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    1099 Build A Binary Search Tree

    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.

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    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
    
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    Sample Output:

    58 25 82 11 38 67 45 73 42
    
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    题意

    二叉搜索树 (BST) 递归定义为具有以下属性的二叉树:

    • 若它的左子树不空,则左子树上所有结点的值均小于它的根结点的值
    • 若它的右子树不空,则右子树上所有结点的值均大于或等于它的根结点的值
    • 它的左、右子树也分别为二叉搜索树

    给定二叉树的具体结构以及一系列不同的整数,只有一种方法可以将这些数填充到树中,以使结果树满足 BST 的定义。

    请你输出结果树的层序遍历

    给定一个 N N N ,所有结点范围在 0 ∼ N − 1 0 \sim N-1 0N1 之间,接下来 N N N 行会输入 0 ∼ N − 1 0 \sim N-1 0N1 的左右孩子的下标,其中 − 1 -1 1 表示空结点。

    最后一行输出 N N N 个结点的值。

    思路
    1. 利用二叉搜索树的性质,其中序遍历的结果一定是有序序列,所以我们将所有结点的值进行排序就可以得到该树中序遍历的结果。
    2. 模拟该树中序遍历的过程,然后将每个值放入对应的位置上。
    3. 输出层次遍历的结果。
    代码
    #include
    using namespace std;
    
    const int N = 110;
    int n;
    int l[N], r[N];	//用哈希表来存储每个结点左孩子的下标
    int a[N], w[N];
    int q[N];
    
    //模拟中序遍历
    void dfs(int u, int& k)
    {
        if (u == -1)  return;
    
        dfs(l[u], k);
        w[u] = a[k++];
        dfs(r[u], k);
    }
    
    //输出层次遍历的结果
    void bfs()
    {
        int hh = 0, tt = 0;
        q[0] = 0;
    
        while (hh <= tt)
        {
            int t = q[hh++];
            if (l[t] != -1)    q[++tt] = l[t];
            if (r[t] != -1)    q[++tt] = r[t];
        }
    
        cout << w[q[0]];
        for (int i = 1; i < n; i++)    cout << " " << w[q[i]];
        cout << endl;
    }
    
    int main()
    {
        cin >> n;
        for (int i = 0; i < n; i++)    cin >> l[i] >> r[i];
        for (int i = 0; i < n; i++)    cin >> a[i];
    
        sort(a, a + n);    //获得中序遍历结果
    
        int k = 0;
        dfs(0, k);
        bfs();
    
        return 0;
    }
    
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  • 原文地址:https://blog.csdn.net/Newin2020/article/details/127079601