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Print nodes of a Binary Search Tree in Top Level Order and Reversed Bottom Level Order alternately

  • Difficulty Level : Expert
  • Last Updated : 22 Sep, 2020

Given a Binary Search Tree, the task is to print the nodes of the BST in the following order:

  • If the BST contains levels numbered from 1 to N then, the printing order is level 1, level N, level 2, level N – 1, and so on.
  • The top-level order (1, 2, …) nodes are printed from left to right, while the bottom level order (N, N-1, …) nodes are printed from right to left.

Examples:

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

Approach: To solve the problem, the idea is to store the nodes of BST in ascending and descending order of levels and node values and print all the nodes of the same level alternatively between ascending and descending order. Follow the steps below to solve the problem:



  • Initialize a Min Heap and a Max Heap to store the nodes in ascending and descending order of level and node values respectively.
  • Perform a level order traversal on the given BST to store the nodes in the respective priority queues.
  • Print all the nodes of each level one by one from the Min Heap followed by the Max Heap alternately.
  • If any level in the Min Heap or Max Heap is found to be already printed, skip to the next level.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
  
#include <bits/stdc++.h>
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<pair<int, int> > great;
  
    // Stores the nodes in ascending order
    // of the level and node values
  
    priority_queue<pair<int, int>,
                   vector<pair<int, int> >,
                   greater<pair<int, int> > >
        small;
  
    // Initialize a stack for
    // level order traversal
    stack<pair<node*, int> > 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<int> 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;
}
Output:
25 48 38 28 12 5 20 36 40 30 22 10

Time Complexity: O(V log(V)), where V denotes the number of vertices in the given Binary Tree
Auxiliary Space: O(V)

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