Given a Binary Tree, the task is for each level is to print the total number of nodes from all lower levels which are less than or equal to every node present at that level.
Examples:
Input: Below is the given tree:
4
/ \
3 5
/ \ / \
10 2 3 1Output: 4 3 0
Explanation:
Nodes in level 1 has 4 nodes as (3) in level 2 and (2, 3, 1) in level 3.
Nodes in level 2 has 3 nodes as (2, 3, 1) in level 3.
Nodes in level 3 does not have any level left below it.Input: Below is the given tree:
4
/ \
7 9
/ / \
1 3 1
Output: 3 3 0
Approach: Follow the steps below to solve the problem:
- Calculate the minimum value at every level using Level Order Traversal.
- Perform Post Order Traversal on the tree and check for every node whether nodes computed in step 1 are greater than or equal to the node. If found to be true, increment count by one at that level, providing that particular level has the node present in the level below it whose value is less than or equal to all the nodes present at that level.
- Print the final array which gives the number of nodes for that level.
Below is the implementation of the above approach:
// C++ program of the // above approach #include <bits/stdc++.h> using namespace std;
// Stores the nodes to be deleted unordered_map< int , bool > mp;
// Structure of a Tree node struct Node {
int key;
struct Node *left, *right;
}; // Function to create a new node Node* newNode( int key)
{ Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return (temp);
} // Function to find the min value // of node for each level void calculateMin(Node* root,
vector< int >& levelMin)
{ queue<Node*> qt;
qt.push(root);
// Count is used to differentiate
// each level of the tree
int count = 1;
int min_v = INT_MAX;
while (!qt.empty()) {
Node* temp = qt.front();
min_v = min(min_v, temp->key);
qt.pop();
if (temp->left) {
qt.push(temp->left);
}
if (temp->right) {
qt.push(temp->right);
}
count--;
if (count == 0) {
levelMin.push_back(min_v);
min_v = INT_MAX;
count = qt.size();
}
}
} // Function to check whether the nodes in // the level below it are smaller // by performing post order traversal void findNodes(Node* root, vector< int >& levelMin,
vector< int >& levelResult, int level)
{ if (root == NULL)
return ;
// Traverse the left subtree
findNodes(root->left, levelMin,
levelResult, level + 1);
// Traverse right subtree
findNodes(root->right, levelMin,
levelResult, level + 1);
// Check from minimum values
// computed at each level
for ( int i = 0; i < level; i++) {
if (root->key <= levelMin[i]) {
levelResult[i] += 1;
}
}
} // Function to print count of // nodes from all lower levels // having values less than the // the nodes in the current level void printNodes(Node* root)
{ vector< int > levelMin;
calculateMin(root, levelMin);
// Stores the number of levels
int numLevels = levelMin.size();
// Stores the required count
// of nodes for each level
vector< int > levelResult(numLevels, 0);
findNodes(root, levelMin, levelResult, 0);
for ( int i = 0; i < numLevels; i++) {
cout << levelResult[i] << " " ;
}
} // Driver Code int main()
{ /*
4
/ \
3 5
/ \ / \
10 2 3 1
*/
Node* root = newNode(4);
root->left = newNode(3);
root->right = newNode(5);
root->right->left = newNode(3);
root->right->right = newNode(1);
root->left->left = newNode(10);
root->left->right = newNode(2);
printNodes(root);
} |
// Java program of the // above approach import java.util.*;
class GFG{
// Stores the nodes to be deleted static Map<Integer,
Boolean> mp = new HashMap<Integer,
Boolean>();
// Structure of a Tree node
static class Node
{ int key;
Node left, right;
}; // Function to create a new node static Node newNode( int key)
{ Node temp = new Node();
temp.key = key;
temp.left = temp.right = null ;
return (temp);
} // Function to find the min value // of node for each level static void calculateMin(Node root,
Vector<Integer> levelMin)
{ Queue<Node> qt = new LinkedList<>();
qt.add(root);
// Count is used to differentiate
// each level of the tree
int count = 1 ;
int min_v = Integer.MAX_VALUE;
while (!qt.isEmpty())
{
Node temp = qt.peek();
min_v = Math.min(min_v, temp.key);
qt.remove();
if (temp.left != null )
{
qt.add(temp.left);
}
if (temp.right != null )
{
qt.add(temp.right);
}
count--;
if (count == 0 )
{
levelMin.add(min_v);
min_v = Integer.MAX_VALUE;
count = qt.size();
}
}
} // Function to check whether the nodes in // the level below it are smaller // by performing post order traversal static void findNodes(Node root,
Vector<Integer> levelMin,
int []levelResult, int level)
{ if (root == null )
return ;
// Traverse the left subtree
findNodes(root.left, levelMin,
levelResult, level + 1 );
// Traverse right subtree
findNodes(root.right, levelMin,
levelResult, level + 1 );
// Check from minimum values
// computed at each level
for ( int i = 0 ; i < level; i++)
{
if (root.key <= levelMin.get(i))
{
levelResult[i] += 1 ;
}
}
} // Function to print count of // nodes from all lower levels // having values less than the // the nodes in the current level static void printNodes(Node root)
{ Vector<Integer> levelMin =
new Vector<Integer>();
calculateMin(root, levelMin);
// Stores the number of levels
int numLevels = levelMin.size();
// Stores the required count
// of nodes for each level
int []levelResult = new int [numLevels];
findNodes(root, levelMin, levelResult, 0 );
for ( int i = 0 ; i < numLevels; i++)
{
System.out.print(levelResult[i] + " " );
}
} // Driver Code public static void main(String[] args)
{ /*
4
/ \
3 5
/ \ / \
10 2 3 1
*/
Node root = newNode( 4 );
root.left = newNode( 3 );
root.right = newNode( 5 );
root.right.left = newNode( 3 );
root.right.right = newNode( 1 );
root.left.left = newNode( 10 );
root.left.right = newNode( 2 );
printNodes(root);
} } // This code is contributed by 29AjayKumar |
# Python3 program of the # above approach from collections import deque
from sys import maxsize as INT_MAX
# Stores the nodes to be deleted mp = dict ()
# Structure of a Tree node class Node:
def __init__( self , key):
self .key = key
self .left = None
self .right = None
# Function to find the min value # of node for each level def calculateMin(root: Node,
levelMin: list ) - > None :
qt = deque()
qt.append(root)
# Count is used to differentiate
# each level of the tree
count = 1
min_v = INT_MAX
while (qt):
temp = qt.popleft()
min_v = min (min_v, temp.key)
if (temp.left):
qt.append(temp.left)
if (temp.right):
qt.append(temp.right)
count - = 1
if (count = = 0 ):
levelMin.append(min_v)
min_v = INT_MAX
count = len (qt)
# Function to check whether the nodes in # the level below it are smaller # by performing post order traversal def findNodes(root: Node,
levelMin: list ,
levelResult: list ,
level: int ) - > None :
if (root = = None ):
return
# Traverse the left subtree
findNodes(root.left, levelMin,
levelResult, level + 1 )
# Traverse right subtree
findNodes(root.right, levelMin,
levelResult, level + 1 )
# Check from minimum values
# computed at each level
for i in range (level):
if (root.key < = levelMin[i]):
levelResult[i] + = 1
# Function to print count of # nodes from all lower levels # having values less than the # the nodes in the current level def printNodes(root: Node) - > None :
levelMin = []
calculateMin(root, levelMin)
# Stores the number of levels
numLevels = len (levelMin)
# Stores the required count
# of nodes for each level
levelResult = [ 0 ] * numLevels
findNodes(root, levelMin, levelResult, 0 )
for i in range (numLevels):
print (levelResult[i], end = " " )
# Driver Code if __name__ = = "__main__" :
'''
4
/ \
3 5
/ \ / \
10 2 3 1
'''
root = Node( 4 )
root.left = Node( 3 )
root.right = Node( 5 )
root.right.left = Node( 3 )
root.right.right = Node( 1 )
root.left.left = Node( 10 )
root.left.right = Node( 2 )
printNodes(root)
# This code is contributed by sanjeev2552 |
// C# program of the // above approach using System;
using System.Collections.Generic;
class GFG{
// Stores the nodes to be deleted static Dictionary< int ,
Boolean> mp = new Dictionary< int ,
Boolean>();
// Structure of a Tree node class Node
{ public int key;
public Node left, right;
}; // Function to create a new node static Node newNode( int key)
{ Node temp = new Node();
temp.key = key;
temp.left = temp.right = null ;
return (temp);
} // Function to find the min value // of node for each level static void calculateMin(Node root,
List< int > levelMin)
{ Queue<Node> qt = new Queue<Node>();
qt.Enqueue(root);
// Count is used to differentiate
// each level of the tree
int count = 1;
int min_v = int .MaxValue;
while (qt.Count != 0)
{
Node temp = qt.Peek();
min_v = Math.Min(min_v, temp.key);
qt.Dequeue();
if (temp.left != null )
{
qt.Enqueue(temp.left);
}
if (temp.right != null )
{
qt.Enqueue(temp.right);
}
count--;
if (count == 0)
{
levelMin.Add(min_v);
min_v = int .MaxValue;
count = qt.Count;
}
}
} // Function to check whether the nodes in // the level below it are smaller // by performing post order traversal static void findNodes(Node root,
List< int > levelMin,
int []levelResult,
int level)
{ if (root == null )
return ;
// Traverse the left subtree
findNodes(root.left, levelMin,
levelResult, level + 1);
// Traverse right subtree
findNodes(root.right, levelMin,
levelResult, level + 1);
// Check from minimum values
// computed at each level
for ( int i = 0; i < level; i++)
{
if (root.key <= levelMin[i])
{
levelResult[i] += 1;
}
}
} // Function to print count of // nodes from all lower levels // having values less than the // the nodes in the current level static void printNodes(Node root)
{ List< int > levelMin = new List< int >();
calculateMin(root, levelMin);
// Stores the number of levels
int numLevels = levelMin.Count;
// Stores the required count
// of nodes for each level
int []levelResult = new int [numLevels];
findNodes(root, levelMin, levelResult, 0);
for ( int i = 0; i < numLevels; i++)
{
Console.Write(levelResult[i] + " " );
}
} // Driver Code public static void Main(String[] args)
{ /* 4
/ \
3 5
/ \ / \
10 2 3 1
*/
Node root = newNode(4);
root.left = newNode(3);
root.right = newNode(5);
root.right.left = newNode(3);
root.right.right = newNode(1);
root.left.left = newNode(10);
root.left.right = newNode(2);
printNodes(root);
} } // This code is contributed by Amit Katiyar |
<script> // JavaScript program for the above approach
// Structure of a Tree node
class Node
{
constructor(key) {
this .left = null ;
this .right = null ;
this .key = key;
}
}
// Function to create a new node
function newNode(key)
{
let temp = new Node(key);
return (temp);
}
// Function to find the min value
// of node for each level
function calculateMin(root, levelMin)
{
let qt = [];
qt.push(root);
// Count is used to differentiate
// each level of the tree
let count = 1;
let min_v = Number.MAX_VALUE;
while (qt.length > 0)
{
let temp = qt[0];
min_v = Math.min(min_v, temp.key);
qt.shift();
if (temp.left != null )
{
qt.push(temp.left);
}
if (temp.right != null )
{
qt.push(temp.right);
}
count--;
if (count == 0)
{
levelMin.push(min_v);
min_v = Number.MAX_VALUE;
count = qt.length;
}
}
}
// Function to check whether the nodes in
// the level below it are smaller
// by performing post order traversal
function findNodes(root, levelMin, levelResult, level)
{
if (root == null )
return ;
// Traverse the left subtree
findNodes(root.left, levelMin,
levelResult, level + 1);
// Traverse right subtree
findNodes(root.right, levelMin,
levelResult, level + 1);
// Check from minimum values
// computed at each level
for (let i = 0; i < level; i++)
{
if (root.key <= levelMin[i])
{
levelResult[i] += 1;
}
}
}
// Function to print count of
// nodes from all lower levels
// having values less than the
// the nodes in the current level
function printNodes(root)
{
let levelMin = [];
calculateMin(root, levelMin);
// Stores the number of levels
let numLevels = levelMin.length;
// Stores the required count
// of nodes for each level
let levelResult = new Array(numLevels);
levelResult.fill(0);
findNodes(root, levelMin, levelResult, 0);
for (let i = 0; i < levelResult.length; i++)
{
document.write(levelResult[i] + " " );
}
}
/*
4
/ \
3 5
/ \ / \
10 2 3 1
*/
let root = newNode(4);
root.left = newNode(3);
root.right = newNode(5);
root.right.left = newNode(3);
root.right.right = newNode(1);
root.left.left = newNode(10);
root.left.right = newNode(2);
printNodes(root);
</script> |
4 3 0
Time Complexity: O(N2)
Auxiliary Space: O(N)