# Binary Tree Maximum Path Sum LeetCode Solution Share:

Binary Tree Maximum Path Sumpath 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`

Time: O(n)
Space: O(h)

### 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);
}
};
``````

### 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);
}
}
``````

### Python

``````
class Solution:
def maxPathSum(self, root: Optional[TreeNode]) -> int:
ans = -math.inf

def maxPathSumDownFrom(root: Optional[TreeNode]) -> int:
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
``````

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