Count root to leaf paths having exactly K distinct nodes in a Binary Tree

Last Updated : 22 Jun, 2021

Given a Binary Tree consisting of N nodes rooted at 1, an integer K and an array arr[] consisting of values assigned to each node, the task is to count the number of root to leaf paths having exactly K distinct nodes in the given Binary Tree.

Examples:

Input: N = 3, Edges[][] = {{1, 2}, {1, 3}}, arr[] = {3, 3, 2}, K = 2, Below is the given Tree:

Output: 1
Explanation:
There exists only 1 distinct path i.e., Path 1 -> 3 contains 2 distinct nodes.
Hence, the answer is 1.

Input: N = 7, Edges[][] = {{1, 2}, {1, 3}, {2, 4}, {2, 5}, {3, 6}, {3, 7}}, arr[] = {2, 2, 2, 2, 3, 5, 2}, K = 1, Below is the given Tree:

Output: 2
Explanation:
There exists only 2 distinct paths containing 1 distinct node:
1) Paths 1 -> 2 -> 4,
2) Paths 1 -> 3 -> 7
Hence, the answer is 2.

Naive Approach: The simplest approach is to generate all possible paths from the root to the leaf nodes and for each path, check if it contains K distinct nodes or not. Finally, print the count of such paths.

Time Complexity: O(N * H²), where H denotes the height of the tree.
Auxiliary Space: O(N);

Efficient Approach: The idea is to use Preorder Traversal and a Map to count the distinct node in the path from the root to the current node. Follow the below steps to solve the problem:

Initialize a variable distinct_nodes as 0 to store the count of the distinct node from the root to the current node and ans as 0 to store the total distinct root to leaf paths having K distinct node.
Perform Preorder Traversal in the given binary tree and store the count of the distinct node from root to the current node in the map M.
Whenever a node occurs for the first time on a path, increase the count of distinct nodes by 1.
If the count of distinct nodes on a path becomes greater than K return to the parent node of the current node.
Otherwise, continue to visit the children of the current node incrementing the frequency of the current node value by 1.
In the above step, increment ans by 1 if the count of distinct nodes on that root to leaf path is exactly equal to K.
After the above steps print the value of ans as the resultant count.

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 Tree Node
struct Node {
    int key;
    Node *left, *right;
};
 
// Function to create new tree node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Function to count all root to leaf
// paths having K distinct nodes
void findkDistinctNodePaths(
    Node* root, unordered_map<int, int> freq,
    int distinct_nodes, int k, int& ans)
{
    // If current node is null
    if (root == NULL)
        return;
 
    // Update count of distinct nodes
    if (freq[root->key] == 0)
        distinct_nodes++;
 
    // If count > k then return to
    // the parent node
    if (distinct_nodes > k)
        return;
 
    // Update frequency of current node
    freq[root->key]++;
 
    // Go to the left subtree
    findkDistinctNodePaths(root->left,
                           freq,
                           distinct_nodes,
                           k, ans);
 
    // Go to the right subtree
    findkDistinctNodePaths(root->right,
                           freq,
                           distinct_nodes,
                           k, ans);
 
    // If current node is leaf node
    if (root->left == NULL
        && root->right == NULL) {
 
        // If count of distinct node
        // is same as K, increment ans
        if (distinct_nodes == k)
            ans++;
    }
}
 
// Function to find count of root to
// leaf paths having K distinct node
void printkDistinctNodePaths(Node* root,
                             int k)
{
    // Initialize unordered map
    unordered_map<int, int> freq;
 
    // Stores count of distinct node
    int distinct_nodes = 0;
 
    // Stores total count of nodes
    int ans = 0;
 
    // Perform Preorder Traversal
    findkDistinctNodePaths(root, freq,
                           distinct_nodes,
                           k, ans);
 
    // Print the final count
    cout << ans;
}
 
// Driver Code
int main()
{
    /*         2
             /   \
            /     \
           1       3
          / \     /  \
         /   \   /    \
        4     2 -5     3
    */
 
    // Given Binary Tree
    Node* root = newNode(2);
    root->left = newNode(1);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(2);
    root->right->left = newNode(-5);
    root->right->right = newNode(3);
 
    // Given K
    int K = 2;
 
    // Function Call
    printkDistinctNodePaths(root, K);
 
    return 0;
}

Java

// Java program for the 
// above approach
import java.util.*;
class GFG{
 
// Structure of a 
// Tree Node
static class Node 
{
  int key;
  Node left, right;
};
   
static int ans;
   
// Function to create 
// new tree node
static Node newNode(int key)
{
  Node temp = new Node();
  temp.key = key;
  temp.left = temp.right = null;
  return temp;
}
 
// Function to count all root 
// to leaf paths having K 
// distinct nodes
static void findkDistinctNodePaths(Node root, 
                                   HashMap<Integer,
                                           Integer> freq, 
                                   int distinct_nodes, 
                                   int k)
{
  // If current node is null
  if (root == null)
    return;
 
  // Update count of distinct nodes
  if (!freq.containsKey(root.key))
    distinct_nodes++;
 
  // If count > k then return 
  // to the parent node
  if (distinct_nodes > k)
    return;
 
  // Update frequency of 
  // current node
  if(freq.containsKey(root.key))
  {
    freq.put(root.key, 
    freq.get(root.key) + 1);
  }
  else
  {
    freq.put(root.key, 1);
  }
 
  // Go to the left subtree
  findkDistinctNodePaths(root.left, freq,
                         distinct_nodes, k);
 
  // Go to the right subtree
  findkDistinctNodePaths(root.right, freq,
                         distinct_nodes, k);
 
  // If current node is 
  // leaf node
  if (root.left == null && 
      root.right == null) 
  {
    // If count of distinct node
    // is same as K, increment ans
    if (distinct_nodes == k)
      ans++;
  }
}
 
// Function to find count of root to
// leaf paths having K distinct node
static void printkDistinctNodePaths(Node root, 
                                    int k)
{
  // Initialize unordered map
  HashMap<Integer,
          Integer> freq = new HashMap<>();
 
  // Stores count of 
  // distinct node
  int distinct_nodes = 0;
 
  // Stores total 
  // count of nodes
  ans = 0;
 
  // Perform Preorder Traversal
  findkDistinctNodePaths(root, freq,
                         distinct_nodes, k);
 
  // Print the final count
  System.out.print(ans);
}
 
// Driver Code
public static void main(String[] args)
{
  /*           2
             /   \
            /     \
           1       3
          / \     /  \
         /   \   /    \
        4     2 -5     3
    */
 
  // Given Binary Tree
  Node root = newNode(2);
  root.left = newNode(1);
  root.right = newNode(3);
  root.left.left = newNode(4);
  root.left.right = newNode(2);
  root.right.left = newNode(-5);
  root.right.right = newNode(3);
 
  // Given K
  int K = 2;
 
  // Function Call
  printkDistinctNodePaths(root, K);
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program for the above approach
 
# Structure of a Tree Node
class newNode:
     
    def __init__(self, key):
         
        self.key = key
        self.left = None
        self.right = None
 
ans = 0
 
# Function to count all root to leaf
# paths having K distinct nodes
def findkDistinctNodePaths(root, freq, 
                           distinct_nodes, k):
                                
    global ans
     
    # If current node is None
    if (root == None):
        return
 
    # Update count of distinct nodes
    if (root.key not in freq):
        distinct_nodes += 1
 
    # If count > k then return to
    # the parent node
    if (distinct_nodes > k):
        return
 
    # Update frequency of current node
    if (root.key in freq):
        freq[root.key] += 1
    else:
        freq[root.key] = freq.get(root.key, 0) + 1
 
    # Go to the left subtree
    findkDistinctNodePaths(root.left, freq,
                           distinct_nodes, k)
 
    # Go to the right subtree
    findkDistinctNodePaths(root.right, freq,
                           distinct_nodes, k)
 
    # If current node is leaf node
    if (root.left == None and
       root.right == None):
         
        # If count of distinct node
        # is same as K, increment ans
        if (distinct_nodes == k):
            ans += 1
 
# Function to find count of root to
# leaf paths having K distinct node
def printkDistinctNodePaths(root, k):
     
    global ans
     
    # Initialize unordered map
    freq = {}
 
    # Stores count of distinct node
    distinct_nodes = 0
 
    # Perform Preorder Traversal
    findkDistinctNodePaths(root, freq,
                           distinct_nodes, k)
 
    # Print the final count
    print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    '''        2
             /   \
            /     \
           1       3
          / \     /  \
         /   \   /    \
        4     2 -5     3
    '''
 
    # Given Binary Tree
    root = newNode(2)
    root.left = newNode(1)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(2)
    root.right.left = newNode(-5)
    root.right.right = newNode(3)
 
    # Given K
    K = 2
 
    # Function Call
    printkDistinctNodePaths(root, K)
     
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the 
// above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Structure of a 
// Tree Node
public class Node 
{
  public int key;
  public Node left, right;
};
   
static int ans;
   
// Function to create 
// new tree node
static Node newNode(int key)
{
  Node temp = new Node();
  temp.key = key;
  temp.left = temp.right = null;
  return temp;
}
   
// Function to count all root 
// to leaf paths having K 
// distinct nodes
static void findkDistinctNodePaths(Node root, 
                                   Dictionary<int, int> freq, 
                                   int distinct_nodes, 
                                   int k)
{
   
  // If current node is null
  if (root == null)
    return;
 
  // Update count of distinct nodes
  if (!freq.ContainsKey(root.key))
    distinct_nodes++;
 
  // If count > k then return 
  // to the parent node
  if (distinct_nodes > k)
    return;
 
  // Update frequency of 
  // current node
  if (freq.ContainsKey(root.key))
  {
    freq[root.key] = freq[root.key] + 1;
  }
  else
  {
    freq.Add(root.key, 1);
  }
 
  // Go to the left subtree
  findkDistinctNodePaths(root.left, freq,
                         distinct_nodes, k);
 
  // Go to the right subtree
  findkDistinctNodePaths(root.right, freq,
                         distinct_nodes, k);
 
  // If current node is 
  // leaf node
  if (root.left == null && 
      root.right == null) 
  {
     
    // If count of distinct node
    // is same as K, increment ans
    if (distinct_nodes == k)
      ans++;
  }
}
 
// Function to find count of root to
// leaf paths having K distinct node
static void printkDistinctNodePaths(Node root, 
                                    int k)
{
   
  // Initialize unordered map
  Dictionary<int,
             int> freq = new Dictionary<int,
                                        int>();
   
  // Stores count of 
  // distinct node
  int distinct_nodes = 0;
 
  // Stores total 
  // count of nodes
  ans = 0;
 
  // Perform Preorder Traversal
  findkDistinctNodePaths(root, freq,
                         distinct_nodes, k);
 
  // Print the readonly count
  Console.Write(ans);
}
 
// Driver Code
public static void Main(String[] args)
{
  /*           2
             /   \
            /     \
           1       3
          / \     /  \
         /   \   /    \
        4     2 -5     3
    */
 
  // Given Binary Tree
  Node root = newNode(2);
  root.left = newNode(1);
  root.right = newNode(3);
  root.left.left = newNode(4);
  root.left.right = newNode(2);
  root.right.left = newNode(-5);
  root.right.right = newNode(3);
 
  // Given K
  int K = 2;
 
  // Function Call
  printkDistinctNodePaths(root, K);
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
// Javascript program for the above approach
 
// Structure of tree node
class Node
{
    constructor(key)
    {
        this.left = null;
        this.right = null;
        this.key = key;
    }
}
 
let ans;
 
// Function to create
// new tree node
function newNode(key)
{
    let temp = new Node(key);
    return temp;
}
 
// Function to count all root
// to leaf paths having K
// distinct nodes
function findkDistinctNodePaths(root, freq, 
                                distinct_nodes, k)
{
     
    // If current node is null
    if (root == null)
        return;
     
    // Update count of distinct nodes
    if (!freq.has(root.key))
        distinct_nodes++;
     
    // If count > k then return
    // to the parent node
    if (distinct_nodes > k)
        return;
     
    // Update frequency of
    // current node
    if (freq.has(root.key))
    {
        freq.set(root.key,
        freq.get(root.key) + 1);
    }
    else
    {
        freq.set(root.key, 1);
    }
     
    // Go to the left subtree
    findkDistinctNodePaths(root.left, freq,
                           distinct_nodes, k);
     
    // Go to the right subtree
    findkDistinctNodePaths(root.right, freq,
                           distinct_nodes, k);
     
    // If current node is
    // leaf node
    if (root.left == null &&
        root.right == null)
    {
         
        // If count of distinct node
        // is same as K, increment ans
        if (distinct_nodes == k)
            ans++;
    }
}
 
// Function to find count of root to
// leaf paths having K distinct node
function printkDistinctNodePaths(root, k)
{
     
    // Initialize unordered map
    let freq = new Map();
     
    // Stores count of
    // distinct node
    let distinct_nodes = 0;
     
    // Stores total
    // count of nodes
    ans = 0;
     
    // Perform Preorder Traversal
    findkDistinctNodePaths(root, freq, 
                           distinct_nodes, k);
     
    // Print the final count
    document.write(ans);
}
 
// Driver code
/*         2
         /   \
        /     \
       1       3
      / \     /  \
     /   \   /    \
    4     2 -5     3
*/
 
// Given Binary Tree
let root = newNode(2);
root.left = newNode(1);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(2);
root.right.left = newNode(-5);
root.right.right = newNode(3);
 
// Given K
let K = 2;
 
// Function Call
printkDistinctNodePaths(root, K);
 
// This code is contributed by suresh07
 
</script>

Output:

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

Root to leaf paths having equal lengths in a Binary Tree

kundudinesh007

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Article Tags :

Practice Tags :

Count root to leaf paths having exactly K distinct nodes in a Binary Tree

C++

Java

Python3

C#

Javascript

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