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Count root to leaf paths having exactly K distinct nodes in a Binary Tree
  • Difficulty Level : Hard
  • Last Updated : 03 Nov, 2020

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
Output: 
2










 

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

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