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124. Binary Tree Maximum Path Sum 👍

  • Time: $O(n)$
  • Space: $O(h)$
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class Solution {
 public:
  int maxPathSum(TreeNode* root) {
    int ans = INT_MIN;
    maxPathSumDownFrom(root, ans);
    return ans;
  }

 private:
  // Returns the maximum path sum starting from the current root, where
  // root->val is always included.
  int maxPathSumDownFrom(TreeNode* root, int& ans) {
    if (root == nullptr)
      return 0;

    const int l = max(0, maxPathSumDownFrom(root->left, ans));
    const int r = max(0, maxPathSumDownFrom(root->right, ans));
    ans = max(ans, root->val + l + r);
    return root->val + max(l, r);
  }
};

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class Solution {
  public int maxPathSum(TreeNode root) {
    maxPathSumDownFrom(root);
    return ans;
  }

  private int ans = Integer.MIN_VALUE;

  // Returns the maximum path sum starting from the current root, where
  // root.val is always included.
  private int maxPathSumDownFrom(TreeNode root) {
    if (root == null)
      return 0;

    final int l = Math.max(maxPathSumDownFrom(root.left), 0);
    final int r = Math.max(maxPathSumDownFrom(root.right), 0);
    ans = Math.max(ans, root.val + l + r);
    return root.val + Math.max(l, r);
  }
}

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class Solution:
  def maxPathSum(self, root: TreeNode | None) -> int:
    ans = -math.inf

    def maxPathSumDownFrom(root: TreeNode | None) -> int:
      """
      Returns the maximum path sum starting from the current root, where
      root.val is always included.
      """
      nonlocal ans
      if not root:
        return 0

      l = max(0, maxPathSumDownFrom(root.left))
      r = max(0, maxPathSumDownFrom(root.right))
      ans = max(ans, root.val + l + r)
      return root.val + max(l, r)

    maxPathSumDownFrom(root)
    return ans