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Print the nodes corresponding to the level value for each level of a Binary Tree
  • Difficulty Level : Hard
  • Last Updated : 12 Oct, 2020

Given a Binary Tree, the task for each level L is to print the Lth node of the tree. If the Lth node is not present for any level, print -1.

Note: Consider the root node to be at the level 1 of the binary tree.

Examples:

Input: Below is the given Tree:

Output:
Level 1: 1
Level 2: 3
Level 3: 6
Level 4: 11
Explanation:
For the first level, the 1st node is 1. 
For the second level, the 2nd node is 3.
For the third level, the 3rd node is 6.
For the fourth level, the 4th node is 11.

Input: Below is the given Tree:

Output:
Level 1: 1
Level 2: 3
Level 3: 6
Level 4: -1
Explanation:
For the first level, the 1st node is 1. 
For the second level, the 2nd node is 3.
For the third level, the 3rd node is 6.
For the fourth level, the 4th node is not available. Hence, print -1.



Approach: To solve this problem the idea is to use Multimap. Follow the steps below to solve the problem:

  1. Traverse the given tree and store the level of each node and the node’s value in the Multimap.
  2. The levels of the nodes are considered as the key of the multimap. Keep track of the maximum level of the Binary Tree(say L).
  3. Now, iterate the Multimap over the range [1, L] and perform the following operations:
    • For each level L, traverse till the Lth node of that level check if it exists or not. If found to be existing, print the value of that node.
    • Otherwise, print “-1” and proceed 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;
  
// Stores the level of the node and
// its value at the max level of BT
multimap<int, int> m;
  
// Stores the maximum level
int maxlevel = 0;
  
// Structure of Binary Tree
struct node {
  
    int data;
    struct node* left;
    struct node* right;
};
  
// Function to insert the node in
// the Binary Tree
struct node* newnode(int d)
{
    struct node* temp
        = (struct node*)malloc(
            sizeof(struct node));
    temp->data = d;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
  
// Function to find node of Nth level
void findNode(struct node* root, int level)
{
    // If root exists
    if (root) {
  
        // Traverse left subtree
        findNode(root->left, level + 1);
  
        // Insert the node's level and
        // its value into the multimap
        m.insert({ level, root->data });
  
        // Update the maximum level
        maxlevel = max(maxlevel, level);
  
        // Traverse the right subtree
        findNode(root->right, level + 1);
    }
}
  
// Function to print the L-th node at
// L-th level of the Binary Tree
void printNode(struct node* root, int level)
{
    // Function Call
    findNode(root, level);
  
    // Iterator for traversing map
    multimap<int, int>::iterator it;
  
    // Iterate all the levels
    for (int i = 0; i <= maxlevel; i++) {
  
        // Print the current level
        cout << "Level " << i + 1 << ": ";
  
        it = m.find(i);
        int flag = 0;
  
        // Iterate upto i-th node of the
        // i-th level
        for (int j = 0; j < i; j++) {
  
            it++;
  
            // If end of the level
            // is reached
            if (it == m.end()) {
                flag = 1;
                break;
            }
        }
  
        // If i-th node does not exist
        // in the i-th level
        if (flag == 1 || it->first != i) {
            cout << "-1" << endl;
        }
  
        // Otherwise
        else {
  
            // Print the i-th node
            cout << it->second << endl;
        }
    }
}
  
// Driver code
int main()
{
    // Construct the Binary Tree
    struct node* root = newnode(1);
    root->left = newnode(2);
    root->right = newnode(3);
  
    root->left->left = newnode(4);
    root->left->right = newnode(5);
  
    root->left->right->left = newnode(11);
    root->left->right->right = newnode(12);
    root->left->left->left = newnode(9);
    root->left->left->right = newnode(10);
    root->right->left = newnode(6);
    root->right->right = newnode(7);
    root->right->right->left = newnode(13);
    root->right->right->right = newnode(14);
  
    // Function Call
    printNode(root, 0);
}
Output:
Level 1: 1
Level 2: 3
Level 3: 6
Level 4: 12

Time Complexity: O(N), where N is the number of nodes in the Binary Tree.
Auxiliary Space: O(N)

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