# Binary Tree Maximum Path Sum | problem No. 124 | LeetCode

Gowthaman Ravi
·Feb 7, 2021·

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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 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

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int maxPathSum(TreeNode* root) {
int res= INT_MIN;
maxSumPath(root, res);
return res;
}

int maxSumPath(TreeNode* root,int& res){
if(root==NULL)
return 0;
int lsum = maxSumPath(root->left,res);
int rsum = maxSumPath(root->right,res);

int max_single = max(max(lsum,rsum)+root->val, root->val);

int max_top = max(max_single, lsum + rsum + root->val);

res = max(res, max_top);

return max_single;
}
};
``````