[Solved] A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root. The path sum of a path is the sum of the node’s values in the path. Given the root of a binary tree, return the maximum path sum of any non-empty path.

Question

path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node’s values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

Example 1:

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2:

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

Constraints:

  • The number of nodes in the tree is in the range [1, 3 * 104].
  • -1000 <= Node.val <= 1000

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
max_sum = -2**31
def DFS(root):
    global max_sum
    if not root: return -2**31
    l = DFS(root.left)
    r = DFS(root.right)
    v = root.val
    max_sum = max(v,v+l,v+r,v+l+r,max_sum)
    return max(l+v,r+v,v)

class Solution:
    def maxPathSum(self, root: TreeNode) -> int:
        global max_sum
        max_sum = -2**31
        DFS(root)
        return max_sum
Abhishek Sharma
Abhishek Sharma

Started my Data Science journey in my 2nd year of college and since then continuously into it because of the magical powers of ML and continuously doing projects in almost every domain of AI like ML, DL, CV, NLP.

Articles: 520

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