In this post, we are going to solve the 124. Binary Tree Maximum Path Sum problem of Leetcode. This problem 124. Binary Tree Maximum Path Sum is a Leetcode hard level problem. Let’s see the code, 124. Binary Tree Maximum Path Sum – Leetcode Solution.

Problem

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

Now, let’s see the code of 124. Binary Tree Maximum Path Sum – Leetcode Solution.

Binary Tree Maximum Path Sum – Leetcode Solution

124. Binary Tree Maximum Path Sum – Solution in Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
    int maxSum=Integer.MIN_VALUE;
    
    public int sum(TreeNode root){
        if(root == null) return 0;
        
        int leftSum = sum(root.left);
        int rightSum = sum(root.right);
        
        int maxInLeftRight = Math.max(leftSum,rightSum);
        int maxInAll = Math.max(root.val,maxInLeftRight+root.val);
        
        int max = Math.max(maxInAll,root.val+leftSum+rightSum);
        
        maxSum = Math.max(maxSum,max);
        return maxInAll;
    }
 
    public int maxPathSum(TreeNode root) {
        sum(root);
        return maxSum;
    }
}

124. Binary Tree Maximum Path Sum – Solution in C++

/**
 * 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 solve(TreeNode* root,int &res)
        {
            // Base Case 
            if(root==NULL) return NULL;
            int ls = solve(root->left,res);
            int rs = solve(root->right,res);
            int temp = max(max(ls,rs)+root->val,root->val);
            int ans = max(temp,ls+rs+root->val);
            res = max(res,ans);
            return temp;
        }
    int maxPathSum(TreeNode* root) {
        int res = INT_MIN;
        solve(root,res);
        return res;
    }
};

124. Binary Tree Maximum Path Sum – Solution in Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution:
    def maxPathSum(self, root: TreeNode) -> int:
        self.max_sum = float('-inf')
        self.dfs(root)
        return self.max_sum
    
    def dfs(self, node):
        if not node: return 0
        
        leftST_sum = max(0, self.dfs(node.left))
        rightST_sum = max(0, self.dfs(node.right))

        self.max_sum = max(self.max_sum, leftST_sum + rightST_sum + node.val)
        
        return max(leftST_sum, rightST_sum) + node.val
        

Note: This problem 124. Binary Tree Maximum Path Sum is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.