Binary Tree Maximum Path Sum

Leetcode

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

Solution

/**
 * 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 maximumPathSum = INT_MIN;
    int dfs(TreeNode* node) {
        if (node == nullptr) {
            return 0;
        }
        int pathSumLeft = dfs(node->left);
        int pathSumRight = dfs(node->right);
        int pathSumAtNode = node->val + pathSumLeft + pathSumRight;
        maximumPathSum = max(maximumPathSum, pathSumAtNode);
        // printf("Val: %d, path-left: %d, path-right: %d, max-path: %d\n",
        //         node->val, pathSumLeft, pathSumRight, maximumPathSum);
        return max(0, max(pathSumLeft + node->val, pathSumRight + node->val));
    }

    int maxPathSum(TreeNode* root) {
        dfs(root);
        return maximumPathSum;
    }
};