Tutorialspoint.dev

Root to leaf path with maximum distinct nodes

Given a Binary Tree, find count of distinct nodes in a root to leaf path with maximum distinct nodes.
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 largestUinquePathUtil(Node* node, unordered_map<int, int> m)
{
    if (!node)
        return m.size();
  
    // put this node into hash
    m[node->data]++;
  
    int max_path = max(largestUinquePathUtil(node->left, m),
                       largestUinquePathUtil(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 largestUinquePath(Node* node)
{
    if (!node)
        return 0;
  
    // hash that store all node value
    unordered_map<int, int> hash;
  
    // return max length unique value path
    return largestUinquePathUtil(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 << largestUinquePath(root) << endl;
  
    return 0;
}

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 largestUinquePathUtil(node, m):
if (not node):
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(largestUinquePathUtil(node.left, m),
largestUinquePathUtil(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 largestUinquePath(node):
if (not node):
return 0

# hash that store all node value
Hash = {}

# return max length unique value path
return largestUinquePathUtil(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(largestUinquePath(root))

# This code is contributed by PranchalK

4

Time Complexity :O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter