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binary tree maximum path sum solution
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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// Time Complexity: O(n)
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// Space Complexity: O(h) where h is the height of the tree
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function maxPathSum(root: TreeNode | null): number {
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let maximumSum = -Infinity;
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const dfs = (node: TreeNode | null) => {
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if (!node) return 0;
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let leftResult = dfs(node.left);
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let rightResult = dfs(node.right);
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leftResult = Math.max(0, leftResult);
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rightResult = Math.max(0, rightResult);
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maximumSum = Math.max(maximumSum, node.val + leftResult + rightResult);
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return node.val + Math.max(leftResult, rightResult);
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};
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dfs(root);
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return maximumSum;
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}

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