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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 Problem** – Binary Tree Maximum Path Sum– LeetCode Problem

Binary Tree Maximum Path Sum– LeetCode Problem

**Problem:**

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:6Explanation: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:42Explanation: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 * 10`

.^{4}] `-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