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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
import { TreeNode } from '../../com_github_leetcode/treenode'
/*
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function maxPathSum(root: TreeNode | null): number {
let max = { sum: Number.NEGATIVE_INFINITY }
dfs(root, max)
return max.sum
}
function dfs(root: TreeNode | null, max) {
if (root == null) {
return 0
}
let l = Math.max(0, dfs(root.left, max))
let r = Math.max(0, dfs(root.right, max))
max.sum = Math.max(max.sum, l + r + root.val)
return root.val + Math.max(l, r)
}
export { maxPathSum }