Print the middle nodes of each level of a Binary Tree

Given a Binary Tree, the task is to print the middle nodes of each level of a binary tree. Considering M to be the number of nodes at any level, print (M/2)^th node if M is odd. Otherwise, print (M/2)^th node and ((M/2) + 1)^th node.

Examples:

Input: Below is the given Tree: 

Output:
1
2 3
5 10
11 6
7 9
Explanation:
The mid nodes of each level is:
Level 0 – 1  
Level 1 – 2 and 3
Level 2 – 5 and 10
Level 3 – 11 and 6
Level 4 – 7 and 9

Input: Below is the given Tree: 

Output:
1
2 3
5
8 9
11
12 13
15 14
Explanation:
The mid nodes of each level is:
Level 0 – 1
Level 1 – 2 and 3
Level 2 – 5
Level 3 – 8 and 9
Level 4 – 11
Level 5 – 12 and 13
Level 6 – 15 and 14

Approach: The idea is to perform the DFS Traversal on the given Tree and store all the nodes for each level in a map of vectors. Now traverse the Map and print the middle nodes accordingly. Below are the steps:

1. Initialize a map of vectors M to stores all the nodes corresponding to each level in a vector.
2. Perform DFS Traversal on the given Tree starting with level 0 and recursively call the left and right subtree with increment in the level by 1.
3. Stores all the nodes in the above DFS Traversal corresponding to each level as M[level].push_back(root->data).
4. Now, traverse the Map M and for each level do the following:
   • Find the size(say S) of the vector(say A) associated with each level in map M.
   • If S is odd then simply print the value of A[(S – 1)/2] as the middle (S/2)^th node.
   • Else print the value of A[(S – 1)/2] and A[(S – 1)/2 + 1] as the middle (S/2)^th and ((S/2) + 1)^th node.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Structure Node of Binary Tree
struct node {
    int data;
    struct node* left;
    struct node* right;
};

// Function to create a new node
struct node* newnode(int d)
{
    struct node* temp
        = (struct node*)malloc(
            sizeof(struct node));
    temp->data = d;
    temp->left = NULL;
    temp->right = NULL;

    // Return the created node
    return temp;
}

// Function that performs the DFS
// traversal on Tree to store all the
// nodes at each level in map M
void dfs(node* root, int l,
         map<int, vector<int> >& M)
{
    // Base Case
    if (root == NULL)
        return;

    // Push the current level node
    M[l].push_back(root->data);

    // Left Recursion
    dfs(root->left, l + 1, M);

    // Right Recursion
    dfs(root->right, l + 1, M);
}

// Function that print all the middle
// nodes for each level in Binary Tree
void printMidNodes(node* root)
{

    // Stores all node in each level
    map<int, vector<int> > M;

    // Perform DFS traversal
    dfs(root, 0, M);

    // Traverse the map M
    for (auto& it : M) {

        // Get the size of vector
        int size = it.second.size();

        // For odd number of elements
        if (size & 1) {

            // Print (M/2)th Element
            cout << it.second[(size - 1) / 2]
                 << endl;
        }

        // Otherwise
        else {

            // Print (M/2)th and
            // (M/2 + 1)th Element
            cout << it.second[(size - 1) / 2]
                 << ' '
                 << it.second[(size - 1) / 2 + 1]
                 << endl;
        }
    }
}

// Driver Code
int main()
{
    /*
    Binary tree shown below is:

                                1
                              /   \
                            2      3
                          /   \   /  \
                         4     5 10   8
                              / \
                             11  6
                                / \
                               7   9


    */

    // Given 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(6);
    root->left->right->right->left = newnode(7);
    root->left->right->right->right = newnode(9);
    root->right->left = newnode(10);
    root->right->right = newnode(8);

    // Function Call
    printMidNodes(root);

    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{

static Map<Integer, Vector<Integer> > M;

// Structure Node of
// Binary Tree
static class node
{
  int data;
  node left;
  node right;
  public node() {}
  public node(int data,
              node left,
              node right)
  {
    super();
    this.data = data;
    this.left = left;
    this.right = right;
  }
};

// Function to create a new node
static node newnode(int d)
{
  node temp = new node();
  temp.data = d;
  temp.left = null;
  temp.right = null;

  // Return the created node
  return temp;
}

// Function that performs the DFS
// traversal on Tree to store all the
// nodes at each level in map M
static void dfs(node root, int l)
{
  // Base Case
  if (root == null)
    return;

  // Push the current level node
  if(!M.containsKey(l))
  {
    Vector<Integer> temp = new Vector<Integer>();
    temp.add(root.data);
    M.put(l, temp);
  }
  else
    M.get(l).add(root.data);

  // Left Recursion
  dfs(root.left, l + 1);

  // Right Recursion
  dfs(root.right, l + 1);
}

// Function that print all the middle
// nodes for each level in Binary Tree
static void printMidNodes(node root)
{
  // Stores all node in each level
  M = new HashMap<Integer,
          Vector<Integer> >();

  // Perform DFS traversal
  dfs(root, 0);

  // Traverse the map M
  for (Map.Entry<Integer,
       Vector<Integer>> it : M.entrySet())
  {
    // Get the size of vector
    int size = it.getValue().size();

    // For odd number of elements
    if (size % 2 == 1)
    {
      // Print (M/2)th Element
      System.out.print(it.getValue().get((size - 1) / 2) + "\n");
    }

    // Otherwise
    else
    {
      // Print (M/2)th and
      // (M/2 + 1)th Element
      System.out.print(it.getValue().get((size - 1) / 2) + " " +
                       it.getValue().get(((size - 1) / 2) + 1) + "\n");
    }
  }
}

// Driver Code
public static void main(String[] args)
{
  /*
    Binary tree shown below is:

                                1
                              /   \
                            2      3
                          /   \   /  \
                         4     5 10   8
                              / \
                             11  6
                                / \
                               7   9


    */

  // Given Tree
  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(6);
  root.left.right.right.left = newnode(7);
  root.left.right.right.right = newnode(9);
  root.right.left = newnode(10);
  root.right.right = newnode(8);

  // Function Call
  printMidNodes(root);
}
}

//This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Structure Node of Binary Tree
class node:

    def __init__(self, data):

        self.data = data
        self.left = None
        self.right = None

# Function to create a new node
def newnode(d):

    temp = node(d)

    # Return the created node
    return temp

# Function that performs the DFS
# traversal on Tree to store all the
# nodes at each level in map M
def dfs(root, l, M):

    # Base Case
    if (root == None):
        return

    # Push the current level node
    if l not in M:
        M[l] = []

    M[l].append(root.data)

    # Left Recursion
    dfs(root.left, l + 1, M)

    # Right Recursion
    dfs(root.right, l + 1, M)

# Function that print all the middle
# nodes for each level in Binary Tree
def printMidNodes(root):

    # Stores all node in each level
    M = dict()

    # Perform DFS traversal
    dfs(root, 0, M)

    # Traverse the map M
    for it in M.values():

        # Get the size of vector
        size = len(it)

        # For odd number of elements
        if (size & 1):

            # Print (M/2)th Element
            print(it[(size - 1) // 2])

        # Otherwise
        else:

            # Print (M/2)th and
            # (M/2 + 1)th Element
            print(str(it[(size - 1) // 2]) + ' ' +
                  str(it[(size - 1) // 2 + 1]))

# Driver Code
if __name__=="__main__":

    '''
    Binary tree shown below is:
            1
          /   \
        2      3
      /   \   /  \
     4     5 10   8
          / \
         11  6
            / \
           7   9
    '''

    # Given Tree
    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(6)
    root.left.right.right.left = newnode(7)
    root.left.right.right.right = newnode(9)
    root.right.left = newnode(10)
    root.right.right = newnode(8)

    # Function Call
    printMidNodes(root)

# This code is contributed by rutvik_56

C#

// C# program for
// the above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG{

static Dictionary<int,ArrayList> M;

// Structure Node of
// Binary Tree
class node
{
  public int data;
  public node left;
  public node right;
  public node(int data,
              node left,
              node right)
  {
    this.data = data;
    this.left = left;
    this.right = right;
  }
};

// Function to create a new node
static node newnode(int d)
{
  node temp = new node(d, null, null);

  // Return the created node
  return temp;
}

// Function that performs the DFS
// traversal on Tree to store all the
// nodes at each level in map M
static void dfs(node root, int l)
{
  // Base Case
  if (root == null)
    return;

  // Push the current level node
  if(!M.ContainsKey(l))
  {
    ArrayList temp = new ArrayList();
    temp.Add(root.data);
    M[l] = temp;
  }
  else
    M[l].Add(root.data);

  // Left Recursion
  dfs(root.left, l + 1);

  // Right Recursion
  dfs(root.right, l + 1);
}

// Function that print all the middle
// nodes for each level in Binary Tree
static void printMidNodes(node root)
{
  // Stores all node in each level
  M = new Dictionary<int, ArrayList>();

  // Perform DFS traversal
  dfs(root, 0);

  // Traverse the map M
  foreach (KeyValuePair<int,ArrayList> it in M)
  {
    // Get the size of vector
    int size = it.Value.Count;

    // For odd number of elements
    if (size % 2 == 1)
    {
      // Print (M/2)th Element
      Console.Write(it.Value[(size - 1) / 2] + "\n");
    }

    // Otherwise
    else
    {

      // Print (M/2)th and
      // (M/2 + 1)th Element
      Console.Write(it.Value[(size - 1) / 2] + " " +
                       it.Value[((size - 1) / 2) + 1] + "\n");
    }
  }
}

// Driver Code
public static void Main(string[] args)
{
  /*
    Binary tree shown below is:

                                1
                              /   \
                            2      3
                          /   \   /  \
                         4     5 10   8
                              / \
                             11  6
                                / \
                               7   9


    */

  // Given Tree
  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(6);
  root.left.right.right.left = newnode(7);
  root.left.right.right.right = newnode(9);
  root.right.left = newnode(10);
  root.right.right = newnode(8);

  // Function Call
  printMidNodes(root);
}
}

// This code is contributed by pratham76

Javascript

<script>

// Javascript program for
// the above approach
var M = new Map();

// Structure Node of
// Binary Tree
class node
{
    constructor(data, left, right)
    {
        this.data = data;
        this.left = left;
        this.right = right;
    }
};

// Function to create a new node
function newnode(d)
{
    var temp = new node(d, null, null);

    // Return the created node
    return temp;
}

// Function that performs the DFS
// traversal on Tree to store all the
// nodes at each level in map M
function dfs(root, l)
{

    // Base Case
    if (root == null)
        return;

    // Push the current level node
    if (!M.has(l))
    {
        var temp = [];
        temp.push(root.data);
        M.set(l, temp);
    }
    else
    {
        var temp = M.get(l);
        temp.push(root.data);
        M.set(l, temp);
    }

    // Left Recursion
    dfs(root.left, l + 1);

    // Right Recursion
    dfs(root.right, l + 1);
}

// Function that print all the middle
// nodes for each level in Binary Tree
function printMidNodes(root)
{

    // Stores all node in each level
    M = new Map();

    // Perform DFS traversal
    dfs(root, 0);

    // Traverse the map M
    M.forEach((value,key) => {

        // Get the size of vector
        var size = value.length;

        // For odd number of elements
        if (size % 2 == 1)
        {
            // Print (M/2)th Element
            document.write(value[parseInt(
                (size - 1) / 2)] + "<br>");
        }

        // Otherwise
        else
        {

            // Print (M/2)th and
            // (M/2 + 1)th Element
            document.write(value[parseInt((size - 1) / 2)] + " " +
                          value[parseInt(((size - 1) / 2) + 1)] + "<br>");
        }
    });
}

// Driver Code
/*
  Binary tree shown below is:

          1
        /   \
      2      3
    /   \   /  \
   4     5 10   8
        / \
       11  6
          / \
         7   9


  */

// Given Tree
var 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(6);
root.left.right.right.left = newnode(7);
root.left.right.right.right = newnode(9);
root.right.left = newnode(10);
root.right.right = newnode(8);

// Function Call
printMidNodes(root);

// This code is contributed by noob2000

</script>

Output:

1
2 3
5 10
11 6
7 9

Time Complexity: O(N²)
Auxiliary Space: O(N)