Binary Tree Maximum Path Sum

Description

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

Time Complexity: $O(N)$, Space Complexity: $O(1)$

/**
 * 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 {
    int maxSum = INT_MIN;
public:
    int maxPathSum(TreeNode* root) {
        maxPath(root);
        return maxSum;
    }
 
    int maxPath(TreeNode* node) {
        if(!node) return 0;
        int leftBranchMax = max(maxPath(node->left), 0);
        int rightBranchMax = max(maxPath(node->right), 0);
        maxSum = max(maxSum, node->val + leftBranchMax + rightBranchMax);
        return node->val + max(leftBranchMax,rightBranchMax);
    }
};

Source

• Binary Tree Maximum Path Sum - LeetCode