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Most Frequent Subtree Sum from a given Binary Tree

  • Difficulty Level : Medium
  • Last Updated : 07 Sep, 2021
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Given a Binary Tree, the task is to find the most frequent subtree sum that can be obtained by considering every node of the given tree as the root of the subtree. If more than one such sums exist, then print all of them.

Examples:

Input:
                 5
              /   \
            2   -4 

Output: 2 -4 3
Explanation:
The sum of nodes considering 5 as the root of subtree is 5 + 2 – 4 = 3.
The sum of nodes considering 2 as the root of subtree is 2 = 2.
The sum of nodes considering -4 as the root of subtree is -4 = -4.
Since all the sums have same frequency, print all the sum.

Input:
              4
           /   \
         2    -4 



Output: 2
Explanation:
The sum of nodes considering 4 as the root of subtree is 4 + 2 – 4 = 2.
The sum of nodes considering 2 as the root of subtree is 2 = 2.
The sum of nodes considering -4 as the root of subtree is -4 = -4.
Since, sum 2 has maximum frequency ( = 2). Hence, print the value 2.

Approach: The idea to use the DFS Traversal for the given tree to solve the given problem. Follow the below steps to solve the problem:

  • Create two auxiliary Hash Map M and F where M is a set of integer keys and corresponding lists, and F will store the frequency of each number.
  • Perform the DFS Traversal for the given tree and do the following:
    • If the node is NULL, return 0.
    • Initialize variables left and right that stores the value of the sum of nodes left and right subtree of the current node.
    • Find the sum of currentNode.value + left + right store it in a variable totalSum.
    • Now update the frequency of totalSum in the map F.
    • Update the frequency of the value F[totalSum] as totalSum in the map M.
    • Return the value to totalSum from the current recursive function.
  • After the above steps, print all the element of the list M.rbegin().

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the vector
void printVector(vector<int> v)
{
    // Traverse vector c
    for (int i = 0; i < v.size(); i++) {
        cout << v[i] << " ";
    }
}
 
// TreeNode class
class TreeNode {
public:
    int val;
    TreeNode *left, *right;
 
    // Constructor
    TreeNode(int data)
    {
        val = data;
        left = NULL;
        right = NULL;
    }
};
 
// Function to insert node in the
// binary tree
void insert(TreeNode** root, int val)
{
    // Initialize Queue
    queue<TreeNode*> q;
 
    // Push the node root
    q.push(*root);
 
    // While Q.size() is there
    while (q.size()) {
 
        // Get the front node
        TreeNode* temp = q.front();
        q.pop();
 
        // If left is NULL
        if (!temp->left) {
            if (val)
                temp->left = new TreeNode(val);
            else
                temp->left = new TreeNode(0);
            return;
        }
        else {
            q.push(temp->left);
        }
 
        // If right is NULL
        if (!temp->right) {
            if (val)
                temp->right = new TreeNode(val);
            else
                temp->right = new TreeNode(0);
            return;
        }
        else {
            q.push(temp->right);
        }
    }
}
 
// Function to make the tree from
// given node values
TreeNode* buildTree(vector<int> v)
{
    TreeNode* root = new TreeNode(v[0]);
 
    // Traverse and insert node
    for (int i = 1; i < v.size(); i++) {
        insert(&root, v[i]);
    }
    return root;
}
 
// Utility function to find subtree
// sum with highest frequency of a
// particular node
int findsubTreeSumUtil(
    TreeNode* node, map<int,
                        vector<int> >& mpp,
    map<int, int>& frequency)
{
    if (!node)
        return 0;
 
    // Recur for the left subtree
    int left = findsubTreeSumUtil(
        node->left, mpp, frequency);
 
    // Recur for the right subtree
    int right = findsubTreeSumUtil(
        node->right, mpp, frequency);
 
    // Stores sum of nodes of a subtree
    int totalSum = node->val + left + right;
 
    // Update the frequency
    if (!frequency.count(totalSum)) {
        mpp[1].push_back(totalSum);
        frequency[totalSum] = 1;
    }
    else {
        frequency[totalSum]++;
        mpp[frequency[totalSum]].push_back(totalSum);
    }
 
    // Return the total sum
    return totalSum;
}
 
// Function to find subtree sum with
// highest frequency of given tree
void findsubTreeSum(TreeNode* root)
{
    // Store list of nodes attached to
    // a particular node and frequency
    // of visited nodes
    map<int, vector<int> > mpp;
    map<int, int> frequency;
 
    // Base Case
    if (!root) {
        printVector({});
        return;
    }
 
    // DFS function call
    findsubTreeSumUtil(root, mpp, frequency);
 
    // Print the vector
    printVector(mpp.rbegin()->second);
}
 
// Driver Code
int main()
{
    // Given nodes of the tree
    vector<int> v = { 5, 2, -4 };
 
    // Function call to build the tree
    TreeNode* tree = buildTree(v);
 
    // Function Call
    findsubTreeSum(tree);
 
    return 0;
}

Java




// Java program for the above approach
 
// Importing required classes
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Queue;
import java.util.TreeMap;
 
public class GFG {
    // Function to print the list
    static void printVector(ArrayList<Integer> v)
    {
        // Traverse list
        for (int i = 0; i < v.size(); i++) {
            System.out.print(v.get(i) + " ");
        }
    }
 
    // TreeNode class
    static class TreeNode {
        int val;
        TreeNode left, right;
 
        // Constructor
        TreeNode(int data)
        {
            val = data;
            left = null;
            right = null;
        }
    }
 
    // Function to insert node in the
    // binary tree
    static void insert(TreeNode root, int val)
    {
        // Initialize Queue
        Queue<TreeNode> q = new LinkedList<TreeNode>();
 
        // Push the node root
        q.add(root);
 
        // While queue is not empty
        while (q.isEmpty() == false) {
 
            // Get the front node
            TreeNode temp = q.peek();
            q.poll();
 
            // If left is NULL
            if (temp.left == null) {
                temp.left = new TreeNode(val);
                return;
            }
            else {
                q.add(temp.left);
            }
 
            // If right is NULL
            if (temp.right == null) {
                temp.right = new TreeNode(val);
                return;
            }
            else {
                q.add(temp.right);
            }
        }
    }
 
    // Function to make the tree from
    // given node values
    static TreeNode buildTree(ArrayList<Integer> v)
    {
        TreeNode root = new TreeNode(v.get(0));
 
        // Traverse and insert node
        for (int i = 1; i < v.size(); i++) {
            insert(root, v.get(i));
        }
        return root;
    }
 
    // Utility function to find subtree
    // sum with highest frequency of a
    // particular node
    static int findsubTreeSumUtil(
        TreeNode node,
        TreeMap<Integer, ArrayList<Integer> > mpp,
        HashMap<Integer, Integer> frequency)
    {
        if (node == null)
            return 0;
 
        // Recur for the left subtree
        int left
            = findsubTreeSumUtil(node.left, mpp, frequency);
 
        // Recur for the right subtree
        int right = findsubTreeSumUtil(node.right, mpp,
                                       frequency);
 
        // Stores sum of nodes of a subtree
        int totalSum = node.val + left + right;
 
        // Update the frequency
        // If there is no node with sub-tree sum equal to
        // totalSum It is the first occurrence of totalSum
        // value
        if (frequency.get(totalSum) == null) {
            ArrayList<Integer> temp = mpp.get(1);
            // If there is no sum which has occurred only
            // once
            if (temp == null)
                temp = new ArrayList<Integer>();
            temp.add(totalSum);
            mpp.put(1, temp);
            frequency.put(totalSum, 1);
        }
        // There is a node with sub-tree sum equal to
        // totalSum
        else {
            frequency.put(totalSum,
                          frequency.get(totalSum) + 1);
            // Get list of sum values which has
            // occurrence equal to occurrence of totalSum
            ArrayList<Integer> temp
                = mpp.get(frequency.get(totalSum));
            // If there is no sum which has occurrence
            // equal to occurrence of totalSum
            if (temp == null)
                temp = new ArrayList<Integer>();
            temp.add(totalSum);
            mpp.put(frequency.get(totalSum), temp);
        }
 
        // Return the total sum
        return totalSum;
    }
 
    // Function to find subtree sum with
    // highest frequency of given tree
    static void findsubTreeSum(TreeNode root)
    {
        // Store list of nodes attached to
        // a particular node and frequency
        // of visited nodes
        TreeMap<Integer, ArrayList<Integer> > mpp
            = new TreeMap<Integer, ArrayList<Integer> >();
        HashMap<Integer, Integer> frequency
            = new HashMap<Integer, Integer>();
 
        // Base Case
        if (root == null) {
            return;
        }
 
        // DFS function call
        findsubTreeSumUtil(root, mpp, frequency);
 
        // Print the list of most-frequent subtree sum
        printVector(mpp.lastEntry().getValue());
    }
 
    // Driver Code
    public static void main(String args[])
    {
        // Given nodes of the tree
        ArrayList<Integer> v = new ArrayList<Integer>();
        v.addAll(Arrays.asList(5, 2, -4));
 
        // Function call to build the tree
        TreeNode tree = buildTree(v);
 
        // Function Call
        findsubTreeSum(tree);
    }
}
 
// This code is contributed by jainlovely450
Output: 
2 -4 3

 

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

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