Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree

Problem Statement❯

Given a binary tree, calculate the sum of the nodes on the longest path from the root node down to a leaf node. If there are multiple longest paths with the same length, return the path with the maximum sum. A binary tree is a data structure where each node has at most two children, referred to as the left and right child.

Examples❯

Input Tree	Sum of Longest Path	Description
[1, 2, 3]	4	Longest paths are 1→2 and 1→3 (length = 2); 2 and 3 tie, but 1+3=4 is greater than 1+2=3
[1, 2, null, 3, null, 4]	10	Longest path is 1 → 2 → 3 → 4 with length 4 and sum 10
[10, 20, 30, 40, null, null, 60]	100	Longest path: 10 → 30 → 60, sum = 100; length = 3
[5, 6, 7, 8, 9, null, null]	19	Longest path is 5 → 6 → 9 (length 3), sum = 5 + 6 + 9 = 20; but 5 → 6 → 8 = 19, so 20 is the correct sum
[1]	1	Single node tree, only one path with length 1 and sum 1
[]	0	Empty tree; no nodes, so sum is 0

Solution❯

Case 1: Tree is Empty

If the tree has no nodes, there are no paths to follow. So, the sum of the longest path is 0. This is the base condition for our recursive logic, and it's a necessary check to prevent null pointer errors.

Case 2: Tree has Only Root Node

In this case, the root node itself is the only node and hence the longest path from root to leaf is the node itself. The sum is simply the value of that root node.

Case 3: Tree has Multiple Paths with Different Lengths

We need to follow all paths from the root to the leaves and keep track of two things: the length of the path and the sum along the way. For each leaf node we reach, we evaluate whether the path we followed is longer than any previously seen path. If it is longer, we update the max sum. If it's of equal length, we update the sum only if it's greater than the current max sum. For example, in a tree like:

    10
   /  \
  20   30
 /       \
40        60

The paths 10→20→40 and 10→30→60 are of equal length, but the second one gives a higher sum. So we pick that.

Case 4: Tree has Balanced Lengths but Different Sums

In some cases, both left and right subtrees might have the same depth. In such scenarios, the final answer is the one with the highest sum. The logic must always check for both length and sum to make the correct decision.

Recursive Approach to Track Length and Sum

When implementing this problem recursively, you need to pass both the current sum and the current path length down the tree. On reaching a leaf, compare its path length and path sum against the maximum values tracked globally (typically using a helper function or a class-level variable).

Algorithm Steps❯

If the tree is empty, return 0.
If the current node is a leaf (no left and right children), return its value.
Recursively calculate the longest path sum for the left subtree and the right subtree.
Select the maximum sum between the left and right subtree.
Add the current node's value to the maximum sum and return the result.

Code

Python

Java

JavaScript

C++

Kotlin

Swift

Php

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def longestPathSum(root):
    if not root:
        return 0
    # If leaf node, return its value
    if not root.left and not root.right:
        return root.val
    leftSum = longestPathSum(root.left)
    rightSum = longestPathSum(root.right)
    return root.val + max(leftSum, rightSum)

# Example usage:
if __name__ == '__main__':
    # Construct binary tree:
    #         1
    #        / \
    #       2   3
    #      /   / \
    #     4   5   6
    root = TreeNode(1, TreeNode(2, TreeNode(4)), TreeNode(3, TreeNode(5), TreeNode(6)))
    print("Longest path sum:", longestPathSum(root))

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