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Sum of nodes on the longest path from root to leaf node

  • Difficulty Level : Medium
  • Last Updated : 27 Sep, 2021

Given a binary tree containing n nodes. The problem is to find the sum of all nodes on the longest path from root to leaf node. If two or more paths compete for the longest path, then the path having maximum sum of nodes is being considered.
Examples: 
 

Input : Binary tree:
        4        
       / \       
      2   5      
     / \ / \     
    7  1 2  3    
      /
     6
Output : 13

        4        
       / \       
      2   5      
     / \ / \     
    7  1 2  3 
      /
     6

The highlighted nodes (4, 2, 1, 6) above are 
part of the longest root to leaf path having
sum = (4 + 2 + 1 + 6) = 13

 

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Approach: Recursively find the length and sum of nodes of each root to leaf path and accordingly update the maximum sum.
Algorithm: 
 



sumOfLongRootToLeafPath(root, sum, len, maxLen, maxSum)
    if root == NULL
        if maxLen < len
        maxLen = len
        maxSum = sum
    else if maxLen == len && maxSum is less than sum
        maxSum = sum
        return

    sumOfLongRootToLeafPath(root-left, sum + root-data,
                           len + 1, maxLen, maxSum)
    sumOfLongRootToLeafPath(root-right, sum + root-data,
                           len + 1, maxLen, maxSum)

sumOfLongRootToLeafPathUtil(root)
    if (root == NULL)
        return 0
    
    Declare maxSum = Minimum Integer
    Declare maxLen = 0
    sumOfLongRootToLeafPath(root, 0, 0, maxLen, maxSum)
    return maxSum

 

C++




// C++ implementation to find the sum of nodes
// on the longest path from root to leaf node
#include <bits/stdc++.h>
 
using namespace std;
 
// Node of a binary tree
struct Node {
    int data;
    Node* left, *right;
};
 
// function to get a new node
Node* getNode(int data)
{
    // allocate memory for the node
    Node* newNode = (Node*)malloc(sizeof(Node));
 
    // put in the data
    newNode->data = data;
    newNode->left = newNode->right = NULL;
    return newNode;
}
 
// function to find the sum of nodes on the
// longest path from root to leaf node
void sumOfLongRootToLeafPath(Node* root, int sum,
                      int len, int& maxLen, int& maxSum)
{
    // if true, then we have traversed a
    // root to leaf path
    if (!root) {
        // update maximum length and maximum sum
        // according to the given conditions
        if (maxLen < len) {
            maxLen = len;
            maxSum = sum;
        } else if (maxLen == len && maxSum < sum)
            maxSum = sum;
        return;
    }
 
    // recur for left subtree
    sumOfLongRootToLeafPath(root->left, sum + root->data,
                            len + 1, maxLen, maxSum);
 
    // recur for right subtree
    sumOfLongRootToLeafPath(root->right, sum + root->data,
                            len + 1, maxLen, maxSum);
}
 
// utility function to find the sum of nodes on
// the longest path from root to leaf node
int sumOfLongRootToLeafPathUtil(Node* root)
{
    // if tree is NULL, then sum is 0
    if (!root)
        return 0;
 
    int maxSum = INT_MIN, maxLen = 0;
 
    // finding the maximum sum 'maxSum' for the
    // maximum length root to leaf path
    sumOfLongRootToLeafPath(root, 0, 0, maxLen, maxSum);
 
    // required maximum sum
    return maxSum;
}
 
// Driver program to test above
int main()
{
    // binary tree formation
    Node* root = getNode(4);         /*        4        */
    root->left = getNode(2);         /*       / \       */
    root->right = getNode(5);        /*      2   5      */
    root->left->left = getNode(7);   /*     / \ / \     */
    root->left->right = getNode(1);  /*    7  1 2  3    */
    root->right->left = getNode(2);  /*      /          */
    root->right->right = getNode(3); /*     6           */
    root->left->right->left = getNode(6);
 
    cout << "Sum = "
         << sumOfLongRootToLeafPathUtil(root);
 
    return 0;
}

Java




// Java implementation to find the sum of nodes
// on the longest path from root to leaf node
public class GFG
{                       
    // Node of a binary tree
    static class Node {
        int data;
        Node left, right;
         
        Node(int data){
            this.data = data;
            left = null;
            right = null;
        }
    }
    static int maxLen;
    static int maxSum;
     
    // function to find the sum of nodes on the
    // longest path from root to leaf node
    static void sumOfLongRootToLeafPath(Node root, int sum,
                                         int len)
    {
        // if true, then we have traversed a
        // root to leaf path
        if (root == null) {
            // update maximum length and maximum sum
            // according to the given conditions
            if (maxLen < len) {
                maxLen = len;
                maxSum = sum;
            } else if (maxLen == len && maxSum < sum)
                maxSum = sum;
            return;
        }
         
         
        // recur for left subtree
        sumOfLongRootToLeafPath(root.left, sum + root.data,
                                len + 1);
         
        sumOfLongRootToLeafPath(root.right, sum + root.data,
                                len + 1);
         
    }
      
    // utility function to find the sum of nodes on
    // the longest path from root to leaf node
    static int sumOfLongRootToLeafPathUtil(Node root)
    {
        // if tree is NULL, then sum is 0
        if (root == null)
            return 0;
      
        maxSum = Integer.MIN_VALUE;
        maxLen = 0;
      
        // finding the maximum sum 'maxSum' for the
        // maximum length root to leaf path
        sumOfLongRootToLeafPath(root, 0, 0);
      
        // required maximum sum
        return maxSum;
    }
      
    // Driver program to test above
    public static void main(String args[])
    {
        // binary tree formation
        Node root = new Node(4);         /*        4        */
        root.left = new Node(2);         /*       / \       */
        root.right = new Node(5);        /*      2   5      */
        root.left.left = new Node(7);    /*     / \ / \     */
        root.left.right = new Node(1);   /*    7  1 2  3    */
        root.right.left = new Node(2);   /*      /          */
        root.right.right = new Node(3);  /*     6           */
        root.left.right.left = new Node(6);
      
        System.out.println( "Sum = "
             + sumOfLongRootToLeafPathUtil(root));
    }
}
// This code is contributed by Sumit Ghosh

Python3




# Python3 implementation to find the 
# Sum of nodes on the longest path
# from root to leaf nodes
 
# function to get a new node
class getNode:
    def __init__(self, data):
 
        # put in the data
        self.data = data
        self.left = self.right = None
 
# function to find the Sum of nodes on the
# longest path from root to leaf node
def SumOfLongRootToLeafPath(root, Sum, Len,
                            maxLen, maxSum):
                                 
    # if true, then we have traversed a
    # root to leaf path
    if (not root):
         
        # update maximum Length and maximum Sum
        # according to the given conditions
        if (maxLen[0] < Len):
            maxLen[0] = Len
            maxSum[0] = Sum
        elif (maxLen[0]== Len and
              maxSum[0] < Sum):
            maxSum[0] = Sum
        return
 
    # recur for left subtree
    SumOfLongRootToLeafPath(root.left, Sum + root.data,
                            Len + 1, maxLen, maxSum)
 
    # recur for right subtree
    SumOfLongRootToLeafPath(root.right, Sum + root.data,
                            Len + 1, maxLen, maxSum)
 
# utility function to find the Sum of nodes on
# the longest path from root to leaf node
def SumOfLongRootToLeafPathUtil(root):
     
    # if tree is NULL, then Sum is 0
    if (not root):
        return 0
 
    maxSum = [-999999999999]
    maxLen = [0]
 
    # finding the maximum Sum 'maxSum' for
    # the maximum Length root to leaf path
    SumOfLongRootToLeafPath(root, 0, 0,
                            maxLen, maxSum)
 
    # required maximum Sum
    return maxSum[0]
 
# Driver Code
if __name__ == '__main__':
     
    # binary tree formation
    root = getNode(4)         #     4    
    root.left = getNode(2)         #     / \    
    root.right = getNode(5)     #     2 5    
    root.left.left = getNode(7) #     / \ / \    
    root.left.right = getNode(1) # 7 1 2 3
    root.right.left = getNode(2) #     /        
    root.right.right = getNode(3) #     6        
    root.left.right.left = getNode(6)
 
    print("Sum = ", SumOfLongRootToLeafPathUtil(root))
     
# This code is contributed by PranchalK

C#




using System;
 
// c# implementation to find the sum of nodes
// on the longest path from root to leaf node
public class GFG
{
    // Node of a binary tree
    public class Node
    {
        public int data;
        public Node left, right;
 
        public Node(int data)
        {
            this.data = data;
            left = null;
            right = null;
        }
    }
    public static int maxLen;
    public static int maxSum;
 
    // function to find the sum of nodes on the
    // longest path from root to leaf node
    public static void sumOfLongRootToLeafPath(Node root, int sum, int len)
    {
        // if true, then we have traversed a
        // root to leaf path
        if (root == null)
        {
            // update maximum length and maximum sum
            // according to the given conditions
            if (maxLen < len)
            {
                maxLen = len;
                maxSum = sum;
            }
            else if (maxLen == len && maxSum < sum)
            {
                maxSum = sum;
            }
            return;
        }
 
 
        // recur for left subtree
        sumOfLongRootToLeafPath(root.left, sum + root.data, len + 1);
 
        sumOfLongRootToLeafPath(root.right, sum + root.data, len + 1);
 
    }
 
    // utility function to find the sum of nodes on
    // the longest path from root to leaf node
    public static int sumOfLongRootToLeafPathUtil(Node root)
    {
        // if tree is NULL, then sum is 0
        if (root == null)
        {
            return 0;
        }
 
        maxSum = int.MinValue;
        maxLen = 0;
 
        // finding the maximum sum 'maxSum' for the
        // maximum length root to leaf path
        sumOfLongRootToLeafPath(root, 0, 0);
 
        // required maximum sum
        return maxSum;
    }
 
    // Driver program to test above
    public static void Main(string[] args)
    {
        // binary tree formation
        Node root = new Node(4); //        4
        root.left = new Node(2); //       / \
        root.right = new Node(5); //      2   5
        root.left.left = new Node(7); //     / \ / \
        root.left.right = new Node(1); //    7  1 2  3
        root.right.left = new Node(2); //      /
        root.right.right = new Node(3); //     6
        root.left.right.left = new Node(6);
 
        Console.WriteLine("Sum = " + sumOfLongRootToLeafPathUtil(root));
    }
}
 
  // This code is contributed by Shrikant13

Javascript




<script>
// javascript implementation to find the sum of nodes
// on the longest path from root to leaf node
          
    // Node of a binary tree
     class Node {
            constructor(val) {
                this.data = val;
                this.left = null;
                this.right = null;
            }
        }
    var maxLen;
    var maxSum;
     
    // function to find the sum of nodes on the
    // longest path from root to leaf node
    function sumOfLongRootToLeafPath(root , sum,
                                          len)
    {
        // if true, then we have traversed a
        // root to leaf path
        if (root == null)
        {
         
            // update maximum length and maximum sum
            // according to the given conditions
            if (maxLen < len) {
                maxLen = len;
                maxSum = sum;
            } else if (maxLen == len && maxSum < sum)
                maxSum = sum;
            return;
        }
         
         
        // recur for left subtree
        sumOfLongRootToLeafPath(root.left, sum + root.data,
                                len + 1);
         
        sumOfLongRootToLeafPath(root.right, sum + root.data,
                                len + 1);
         
    }
      
    // utility function to find the sum of nodes on
    // the longest path from root to leaf node
    function sumOfLongRootToLeafPathUtil(root)
    {
     
        // if tree is NULL, then sum is 0
        if (root == null)
            return 0;
      
        maxSum = Number.MIN_VALUE;
        maxLen = 0;
      
        // finding the maximum sum 'maxSum' for the
        // maximum length root to leaf path
        sumOfLongRootToLeafPath(root, 0, 0);
      
        // required maximum sum
        return maxSum;
    }
      
    // Driver program to test above
     
        // binary tree formation
        var root = new Node(4);         /*        4        */
        root.left = new Node(2);         /*       / \       */
        root.right = new Node(5);        /*      2   5      */
        root.left.left = new Node(7);    /*     / \ / \     */
        root.left.right = new Node(1);   /*    7  1 2  3    */
        root.right.left = new Node(2);   /*      /          */
        root.right.right = new Node(3);  /*     6           */
        root.left.right.left = new Node(6);
      
        document.write( "Sum = "
             + sumOfLongRootToLeafPathUtil(root));
 
// This code is contributed by gauravrajput1
</script>
Output
Sum = 13

Time Complexity: O(n) 

Another Approach: Using level order traversal

  1. Create a structure containing the current Node, level and sum in the path.
  2. Push the root element with level 0 and sum as the root’s data.
  3. Pop the front element and update the maximum level sum and maximum level if needed.
  4. Push the left and right nodes if exists.
  5. Do the same for all the nodes in tree.

C++




#include <bits/stdc++.h>
using namespace std;
 
//Building a tree node having left and right pointers set to null initially
struct Node
{
  Node* left;
  Node* right;
  int data;
  //constructor to set the data of the newly created tree node
  Node(int element){
     data = element;
     this->left = nullptr;
     this->right = nullptr;
  }
};
 
int longestPathLeaf(Node* root){
   
  /* structure to store current Node,it's level and sum in the path*/
  struct Element{
    Node* data;
    int level;
    int sum;
  };
   
  /*
    maxSumLevel stores maximum sum so far in the path
    maxLevel stores maximum level so far
  */
  int maxSumLevel = root->data,maxLevel = 0;
 
  /* queue to implement level order traversal */
   
  list<Element> que;
  Element e;
   
  /* Each element variable stores the current Node, it's level, sum in the path */
 
  e.data = root;
  e.level = 0;
  e.sum = root->data;
   
  /* push the root element*/
  que.push_back(e);
   
  /* do level order traversal on the tree*/
  while(!que.empty()){
 
     Element front = que.front();
     Node* curr = front.data;
     que.pop_front();
      
     /* if the level of current front element is greater than the maxLevel so far then update maxSum*/
     if(front.level > maxLevel){
        maxSumLevel = front.sum;
        maxLevel = front.level;
     }
     /* if another path competes then update if the sum is greater than the previous path of same height*/
     else if(front.level == maxLevel && front.sum > maxSumLevel)
        maxSumLevel = front.sum;
 
     /* push the left element if exists*/ 
     if(curr->left){
        e.data = curr->left;
        e.sum = e.data->data;
        e.sum +=  front.sum;
        e.level = front.level+1;
        que.push_back(e);
     }
     /*push the right element if exists*/
     if(curr->right){
        e.data = curr->right;
        e.sum = e.data->data;
        e.sum +=  front.sum;
        e.level = front.level+1;
        que.push_back(e);
     }
  }
 
  /*return the answer*/
  return maxSumLevel;
}
//Helper function
int main() {
   
  Node* root = new Node(4);        
  root->left = new Node(2);        
  root->right = new Node(5);       
  root->left->left = new Node(7);  
  root->left->right = new Node(1); 
  root->right->left = new Node(2);
  root->right->right = new Node(3);
  root->left->right->left = new Node(6);
   
  cout << longestPathLeaf(root) << "\n";
     
  return 0;
}
Output
13

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

Contributed by Manjukrishna

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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