BinaryTreeMaximumPathSum.cpp

/* Problem Statement ->
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: 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
*/

/* Code -> */

/**
 * 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) {}
 * };
 */

#include <iostream>
#include <algorithm>
#include <climits> // Add this include for INT_MIN

using namespace std;

// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int maxSum = INT_MIN;
        calculateMaxPathSum(root, maxSum);
        return maxSum;
    }
    
    int calculateMaxPathSum(TreeNode* node, int& maxSum) {
        if (!node) return 0;
        
        int leftMax = max(0, calculateMaxPathSum(node->left, maxSum));
        int rightMax = max(0, calculateMaxPathSum(node->right, maxSum));
        
        maxSum = max(maxSum, node->val + leftMax + rightMax);
        
        return node->val + max(leftMax, rightMax);
    }
};

int main() {
    // Example 1
    TreeNode* root1 = new TreeNode(1);
    root1->left = new TreeNode(2);
    root1->right = new TreeNode(3);
    Solution solution;
    cout << "Example 1: " << solution.maxPathSum(root1) << endl;

    // Example 2
    TreeNode* root2 = new TreeNode(-10);
    root2->left = new TreeNode(9);
    root2->right = new TreeNode(20);
    root2->right->left = new TreeNode(15);
    root2->right->right = new TreeNode(7);
    cout << "Example 2: " << solution.maxPathSum(root2) << endl;

    return 0;
}