Binary Tree Maximum Path Sum LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Binary Tree Maximum Path Sum in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemBinary Tree Maximum Path Sum– LeetCode Problem

Binary Tree Maximum Path Sum– LeetCode Problem

Problem:

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
Binary Tree Maximum Path Sum– LeetCode Solutions
Binary Tree Maximum Path Sum Solution in C++:
class Solution {
 public:
  int maxPathSum(TreeNode* root) {
    int ans = INT_MIN;

    maxPathSumDownFrom(root, ans);

    return ans;
  }

 private:
  // root->val + 0/1 of its subtrees
  int maxPathSumDownFrom(TreeNode* root, int& ans) {
    if (!root)
      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);
  }
};
Binary Tree Maximum Path Sum Solution in Java:
class Solution {
  public int maxPathSum(TreeNode root) {
    maxPathSumDownFrom(root);
    return ans;
  }

  private int ans = Integer.MIN_VALUE;

  // root->val + 0/1 of its subtrees
  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);
  }
}
Binary Tree Maximum Path Sum Solution in Python:
class Solution:
  def maxPathSum(self, root: Optional[TreeNode]) -> int:
    self.ans = -inf

    def maxPathSumDownFrom(root: Optional[TreeNode]) -> int:
      if not root:
        return 0

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

      return root.val + max(l, r)

    maxPathSumDownFrom(root)
    return self.ans

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