Sum of nodes on the longest path from root to leaf node
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
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 else if (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)
Auxiliary Space: O(h) where h is the height of the binary tree.
Another Approach: Using level order traversal
- Create a structure containing the current Node, level and sum in the path.
- Push the root element with level 0 and sum as the root’s data.
- Pop the front element and update the maximum level sum and maximum level if needed.
- Push the left and right nodes if exists.
- Do the same for all the nodes in tree.
Below is the implementation of the above approach.
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; } |
Java
// Java Code to find sum of nodes on the longest path from // root to leaf node using level order traversal import java.util.*; // Building a tree node having left and right pointers set // to null initially class Main { static class 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 = null ; this .right = null ; } } /* structure to store current Node,it's level and sum in * the path*/ static class Element { Node data; int level; int sum; } public static int longestPathLeaf(Node root) { /* maxSumLevel stores maximum sum so far in the path maxLevel stores maximum level so far */ int maxSumLevel = root.data; int maxLevel = 0 ; /* queue to implement level order traversal */ Queue<Element> que = new LinkedList<>(); Element e = new Element(); /* 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.add(e); /* do level order traversal on the tree*/ while (!que.isEmpty()) { Element front = que.poll(); Node curr = front.data; /* 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 != null ) { e = new Element(); e.data = curr.left; e.sum = e.data.data; e.sum += front.sum; e.level = front.level + 1 ; que.add(e); } /*push the right element if exists*/ if (curr.right != null ) { e = new Element(); e.data = curr.right; e.sum = e.data.data; e.sum += front.sum; e.level = front.level + 1 ; que.add(e); } } /*return the answer*/ return maxSumLevel; } // Helper function public static void main(String[] args) { 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 ); System.out.println(longestPathLeaf(root)); } } // This code is contributed by Tapesh(tapeshdua420) |
Python3
# Python Code to find sum of nodes on the longest path from root to leaf node # using level order traversal # Building a tree node having left and right pointers set to null initially class Node: def __init__( self , element): self .data = element self .left = None self .right = None # To store current Node,it's level and sum in the path class Element: def __init__( self , data, level, sum ): self .data = data self .level = level self . sum = sum class Solution: def longestPathLeaf( self , root): # maxSumLevel stores maximum sum so far in the path # maxLevel stores maximum level so far maxSumLevel = root.data maxLevel = 0 # queue to implement level order traversal que = [] # Each element variable stores the current Node, it's level, sum in the path e = Element(root, 0 , root.data) # push the root element que.append(e) # do level order traversal on the tree while len (que) ! = 0 : front = que[ 0 ] curr = front.data del que[ 0 ] # 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 elif front.level = = maxLevel and front. sum > maxSumLevel: maxSumLevel = front. sum # push the left element if exists if curr.left ! = None : e = Element(curr.left, front.level + 1 , curr.left.data + front. sum ) que.append(e) # push the right element if exists if curr.right ! = None : e = Element(curr.right, front.level + 1 , curr.right.data + front. sum ) que.append(e) # return the answer return maxSumLevel # Helper function if __name__ = = '__main__' : s = Solution() root = Node( 4 ) root.left = Node( 2 ) root.right = Node( 5 ) root.left.left = Node( 7 ) root.left.right = Node( 1 ) root.right.left = Node( 2 ) root.right.right = Node( 3 ) root.left.right.left = Node( 6 ) print (s.longestPathLeaf(root)) # This code is contributed by Tapesh(tapeshdua420) |
C#
// C# program to find sum of nodes on the longest path from // root to leaf node using level order traversal using System; using System.Collections; // Building a tree node having left and right pointers set // to null initially class Node { public Node left; public Node right; public int data; // constructor to set the data of the newly created // tree node public Node( int element) { data = element; this .left = null ; this .right = null ; } } /* structure to store current Node,it's level and sum in * the path*/ class Element { public Node data; public int level; public int sum; } class GFG { public static int longestPathLeaf(Node root) { /* maxSumLevel stores maximum sum so far in the path maxLevel stores maximum level so far */ int maxSumLevel = root.data; int maxLevel = 0; /* queue to implement level order traversal */ Queue que = new Queue(); Element e = new Element(); /* 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.Enqueue(e); /* do level order traversal on the tree*/ while (que.Count != 0) { dynamic front = que.Dequeue(); Node curr = front.data; /* 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 != null ) { e = new Element(); e.data = curr.left; e.sum = e.data.data; e.sum += front.sum; e.level = front.level + 1; que.Enqueue(e); } /*push the right element if exists*/ if (curr.right != null ) { e = new Element(); e.data = curr.right; e.sum = e.data.data; e.sum += front.sum; e.level = front.level + 1; que.Enqueue(e); } } /*return the answer*/ return maxSumLevel; } // Helper function public static void Main( string [] args) { 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); Console.WriteLine(longestPathLeaf(root)); } } // This code is contributed by Tapesh(tapeshdua420) |
Javascript
// Javascript code to find sum of nodes on the longest path from root to leaf node // using level order traversal class Node { constructor(element) { this .data = element; this .left = null ; this .right = null ; } } class Element { constructor(data, level, sum) { this .data = data; this .level = level; this .sum = sum; } } class Solution { longestPathLeaf(root) { // maxSumLevel stores maximum sum so far in the path // maxLevel stores maximum level so far let maxSumLevel = root.data; let maxLevel = 0; // queue to implement level order traversal let que = []; // Each element variable stores the current Node, it's level, sum in the path let e = new Element(root, 0, root.data); // push the root element que.push(e); // do level order traversal on the tree while (que.length !== 0) { let front = que[0]; let curr = front.data; que.shift(); // 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 !== null ) { e = new Element(curr.left, front.level + 1, curr.left.data + front.sum); que.push(e); } // push the right element if exists if (curr.right !== null ) { e = new Element(curr.right, front.level + 1, curr.right.data + front.sum); que.push(e); } } // return the answer return maxSumLevel; } } // Helper function const s = new Solution(); const 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); console.log(s.longestPathLeaf(root)); |
Output
13
Time Complexity: O(N)
Auxiliary Space: O(N)
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