LeetCode-in-All
124. Binary Tree Maximum Path Sum
Hard
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.
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
Solution
; Definition for a binary tree node.
#|
; val : integer?
; left : (or/c tree-node? #f)
; right : (or/c tree-node? #f)
(struct tree-node
(val left right) #:mutable #:transparent)
; constructor
(define (make-tree-node [val 0])
(tree-node val #f #f))
|#
(define max-box (box -inf.0)) ; Rename max to avoid conflict
(define/contract (max-path-sum root)
(-> (or/c tree-node? #f) exact-integer?)
(set-box! max-box -inf.0) ; Reset max-box before calculation
(helper root)
(unbox max-box)) ; Return the stored max value
(define (helper node)
(if (not node)
0
(let* ((left (max 0 (helper (tree-node-left node))))
(right (max 0 (helper (tree-node-right node))))
(current (+ (tree-node-val node) left right)))
(when (> current (unbox max-box)) ; Correctly update max-box
(set-box! max-box current))
(+ (tree-node-val node) (max left right))))) ; Choose the max path