Maximum distinct nodes in a Root to leaf path

Given a Binary Tree, find count of distinct nodes in all the root to leaf paths and print the maximum. 

Examples:

Input :   1
        /    \
       2      3
      / \    / \
     4   5  6   3
             \   \
              8   9 
Output : 4 
The root to leaf path with maximum distinct
nodes is 1-3-6-8.

A simple solution is to explore all root to leaf paths. In every root to leaf path, count distinct nodes and finally return the maximum count.

An efficient solution is to use hashing. We recursively traverse the tree and maintain count of distinct nodes on path from root to current node. We recur for left and right subtrees and finally return maximum of two values.

Below is implementation of above idea 

C++

// C++ program to find count of distinct nodes
// on a path with maximum distinct nodes.
#include <bits/stdc++.h>
using namespace std;

// A node of binary tree
struct Node {
    int data;
    struct Node *left, *right;
};

// A utility function to create a new Binary
// Tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

int largestUniquePathUtil(Node* node, unordered_map<int, int> m)
{
    if (!node)
        return m.size();

    // put this node into hash
    m[node->data]++;

    int max_path = max(largestUniquePathUtil(node->left, m),
                       largestUniquePathUtil(node->right, m));

    // remove current node from path "hash"
    m[node->data]--;

    // if we reached a condition where all duplicate value
    // of current node is deleted
    if (m[node->data] == 0)
        m.erase(node->data);

    return max_path;
}

// A utility function to find long unique value path
int largestUniquePath(Node* node)
{
    if (!node)
        return 0;

    // hash that store all node value
    unordered_map<int, int> hash;

    // return max length unique value path
    return largestUniquePathUtil(node, hash);
}

// Driver program to test above functions
int main()
{
    // Create binary tree shown in above figure
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->right->left->right = newNode(8);
    root->right->right->right = newNode(9);

    cout << largestUniquePath(root) << endl;

    return 0;
}

Java

// Java program to find count of distinct nodes
// on a path with maximum distinct nodes.
import java.util.*;
class GFG
{

// A node of binary tree
static class Node
{
    int data;
    Node left, right;
};

// A utility function to create a new Binary
// Tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}

static int largestUniquePathUtil(Node node, HashMap<Integer,
                                                    Integer> m)
{
    if (node == null)
        return m.size();

    // put this node into hash
    if(m.containsKey(node.data))
    {
        m.put(node.data, m.get(node.data) + 1);
    }
    else
    {
        m.put(node.data, 1);
    }

    int max_path = Math.max(largestUniquePathUtil(node.left, m),
                            largestUniquePathUtil(node.right, m));

    // remove current node from path "hash"
    if(m.containsKey(node.data))
    {
        m.put(node.data, m.get(node.data) - 1);
    }

    // if we reached a condition where all duplicate value
    // of current node is deleted
    if (m.get(node.data) == 0)
        m.remove(node.data);

    return max_path;
}

// A utility function to find long unique value path
static int largestUniquePath(Node node)
{
    if (node == null)
        return 0;

    // hash that store all node value
    HashMap<Integer,
            Integer> hash = new HashMap<Integer,
                                        Integer>();

    // return max length unique value path
    return largestUniquePathUtil(node, hash);
}

// Driver Code
public static void main(String[] args)
{
    // Create binary tree shown in above figure
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
    root.right.right.right = newNode(9);

    System.out.println(largestUniquePath(root));
}
}

// This code is contributed by Princi Singh

Python3

# Python3 program to find count of
# distinct nodes on a path with
# maximum distinct nodes.

# A utility class to create a
# new Binary Tree node
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None

def largestUniquePathUtil(node, m):
    if (not node or node.data in m):
        return len(m)

    # put this node into hash
    if node.data in m:
        m[node.data] += 1
    else:
        m[node.data] = 1

    max_path = max(largestUniquePathUtil(node.left, m),
                   largestUniquePathUtil(node.right, m))

    # remove current node from path "hash"
    m[node.data] -= 1

    # if we reached a condition
    # where all duplicate value
    # of current node is deleted
    if (m[node.data] == 0):
        del m[node.data]

    return max_path

# A utility function to find
# long unique value path
def largestUniquePath(node):
    if (not node):
        return 0

    # hash that store all node value
    Hash = {}

    # return max length unique value path
    return largestUniquePathUtil(node, Hash)

# Driver Code
if __name__ == '__main__':

    # Create binary tree shown
    # in above figure
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.left = newNode(6)
    root.right.right = newNode(7)
    root.right.left.right = newNode(8)
    root.right.right.right = newNode(9)

    print(largestUniquePath(root))

# This code is contributed by PranchalK

C#

// C# program to find count of distinct nodes
// on a path with maximum distinct nodes.
using System;
using System.Collections.Generic;

class GFG
{

// A node of binary tree
public class Node
{
    public int data;
    public Node left, right;
};

// A utility function to create a new Binary
// Tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}

static int largestUniquePathUtil(Node node,
                                 Dictionary<int, int> m)
{
    if (node == null)
        return m.Count;

    // put this node into hash
    if(m.ContainsKey(node.data))
    {
        m[node.data] = m[node.data] + 1;
    }
    else
    {
        m.Add(node.data, 1);
    }

    int max_path = Math.Max(largestUniquePathUtil(node.left, m),
                            largestUniquePathUtil(node.right, m));

    // remove current node from path "hash"
    if(m.ContainsKey(node.data))
    {
        m[node.data] = m[node.data] - 1;
    }

    // if we reached a condition where all
    // duplicate value of current node is deleted
    if (m[node.data] == 0)
        m.Remove(node.data);

    return max_path;
}

// A utility function to find long unique value path
static int largestUniquePath(Node node)
{
    if (node == null)
        return 0;

    // hash that store all node value
    Dictionary<int,
               int> hash = new Dictionary<int,
                                          int>();

    // return max length unique value path
    return largestUniquePathUtil(node, hash);
}

// Driver Code
public static void Main(String[] args)
{

    // Create binary tree shown in above figure
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
    root.right.right.right = newNode(9);

    Console.WriteLine(largestUniquePath(root));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

    // JavaScript program to find count of distinct nodes
    // on a path with maximum distinct nodes.

    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    // A utility function to create a new Binary
    // Tree node
    function newNode(data)
    {
        let temp = new Node(data);
        return temp;
    }

    function largestUniquePathUtil(node, m)
    {
        if (node == null)
            return m.size;

        // put this node into hash
        if(m.has(node.data))
        {
            m.set(node.data, m.get(node.data) + 1);
        }
        else
        {
            m.set(node.data, 1);
        }

        let max_path = Math.max(largestUniquePathUtil(node.left, m),
                                largestUniquePathUtil(node.right, m));

        // remove current node from path "hash"
        if(m.has(node.data))
        {
            m.set(node.data, m.get(node.data) - 1);
        }

        // if we reached a condition where all duplicate value
        // of current node is deleted
        if (m.get(node.data) == 0)
            m.delete(node.data);

        return max_path;
    }

    // A utility function to find long unique value path
    function largestUniquePath(node)
    {
        if (node == null)
            return 0;

        // hash that store all node value
        let hash = new Map();

        // return max length unique value path
        return largestUniquePathUtil(node, hash);
    }

    // Create binary tree shown in above figure
    let root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.right.left.right = newNode(8);
    root.right.right.right = newNode(9);

    document.write(largestUniquePath(root));

</script>

Output

4

Time Complexity: O(n) 

Auxiliary Space: O(n)

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