交替按顶层顺序和底层最低顺序打印二叉搜索树的节点

给定二进制搜索树，任务是按以下顺序打印BST的节点：

• 如果BST包含从1到N的级别，则打印顺序为1级， N级， 2级， N-1级，依此类推。
• 顶层顺序( 1，2 ，…)节点从左到右打印，而底层顺序( N ， N-1 ，…)节点从右到左打印。

例子：

Input:

Output: 27 42 31 19 10 14 35
Explanation:
Level 1 from left to right: 27
Level 3 from right to left: 42 31 19 10
Level 2 from left to right: 14 35

Input:

Output: 25 48 38 28 12 5 20 36 40 30 22 10

方法：解决该问题的方法是将BST的节点按级别和节点值的升序和降序存储，并在升序和降序之间交替打印同一级别的所有节点。请按照以下步骤解决问题：

• 初始化最小堆和最大堆，分别以级别和节点值的升序和降序存储节点。
• 在给定的BST上执行级别顺序遍历，以将节点存储在相应的优先级队列中。
• 从“最小堆”开始依次打印每个级别的所有节点，然后交替打印“最大堆”。
• 如果发现“最小堆”或“最大堆”中的任何级别已被打印，请跳至下一个级别。

下面是上述方法的实现：

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
// Structure of a BST node
struct node {
    int data;
    struct node* left;
    struct node* right;
};
  
// Utility function to create a new BST node
struct node* newnode(int d)
{
    struct node* temp
        = (struct node*)malloc(sizeof(struct node));
    temp->left = NULL;
    temp->right = NULL;
    temp->data = d;
    return temp;
}
  
// Function to print the nodes of a
// BST in Top Level Order and Reversed
// Bottom Level Order alternatively
void printBST(node* root)
{
    // Stores the nodes in descending order
    // of the level and node values
    priority_queue > great;
  
    // Stores the nodes in ascending order
    // of the level and node values
  
    priority_queue,
                   vector >,
                   greater > >
        small;
  
    // Initialize a stack for
    // level order traversal
    stack > st;
  
    // Push the root of BST
    // into the stack
    st.push({ root, 1 });
  
    // Perform Level Order Traversal
    while (!st.empty()) {
  
        // Extract and pop the node
        // from the current level
        node* curr = st.top().first;
  
        // Stores level of current node
        int level = st.top().second;
        st.pop();
  
        // Store in the priority queues
        great.push({ level, curr->data });
        small.push({ level, curr->data });
  
        // Traverse left subtree
        if (curr->left)
            st.push({ curr->left, level + 1 });
  
        // Traverse right subtree
        if (curr->right)
            st.push({ curr->right, level + 1 });
    }
  
    // Stores the levels that are printed
    unordered_set levelsprinted;
  
    // Print the nodes in the required manner
    while (!small.empty() && !great.empty()) {
  
        // Store the top level of traversal
        int toplevel = small.top().first;
  
        // If the level is already printed
        if (levelsprinted.find(toplevel)
            != levelsprinted.end())
            break;
  
        // Otherwise
        else
            levelsprinted.insert(toplevel);
  
        // Print nodes of same level
        while (!small.empty()
               && small.top().first == toplevel) {
            cout << small.top().second << " ";
            small.pop();
        }
  
        // Store the bottom level of traversal
        int bottomlevel = great.top().first;
  
        // If the level is already printed
        if (levelsprinted.find(bottomlevel)
            != levelsprinted.end()) {
            break;
        }
        else {
            levelsprinted.insert(bottomlevel);
        }
  
        // Print the nodes of same level
        while (!great.empty()
               && great.top().first == bottomlevel) {
            cout << great.top().second << " ";
            great.pop();
        }
    }
}
  
// Driver Code
int main()
{
    /*
    Given BST
  
                                25
                              /     \
                            20      36
                           /  \      / \
                          10   22   30 40
                         /  \      /   / \
                        5   12    28  38 48
    */
  
    // Creating the BST
    node* root = newnode(25);
    root->left = newnode(20);
    root->right = newnode(36);
    root->left->left = newnode(10);
    root->left->right = newnode(22);
    root->left->left->left = newnode(5);
    root->left->left->right = newnode(12);
    root->right->left = newnode(30);
    root->right->right = newnode(40);
    root->right->left->left = newnode(28);
    root->right->right->left = newnode(38);
    root->right->right->right = newnode(48);
  
    // Function Call
    printBST(root);
  
    return 0;
}

输出：

25 48 38 28 12 5 20 36 40 30 22 10

时间复杂度： O(V log(V))，其中V表示给定二叉树中的顶点数
辅助空间： O(V)