Last updated on January 17th, 2025 at 01:00 am

Here, we see a Binary Tree Maximum Path Sum LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Depth-First Search, Tree

Companies

Baidu, Microsoft

Level of Question

Hard

Binary Tree Maximum Path Sum LeetCode Solution

1. 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.

2. Coding Pattern Used in Solution

The coding pattern used in the provided code is “Tree Depth First Search (DFS)”. The code traverses the binary tree using a recursive DFS approach to calculate the maximum path sum. This pattern is commonly used when solving problems that require exploring all possible paths in a tree structure.

3. Code Implementation in Different Languages

3.1 Binary Tree Maximum Path Sum C++

class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int maxPath = INT_MIN;
        DFS(root, maxPath);
        return maxPath;
    }
private:
    int DFS(TreeNode* root, int& maxPath) {
        if (root == NULL) {
            return 0;  // Base case: return 0 if the node is NULL
        }
        int lmax = max(DFS(root->left, maxPath), 0);
        int rmax = max(DFS(root->right, maxPath), 0);
        maxPath = max(maxPath, root->val + lmax + rmax);
        return root->val + max(lmax, rmax);
    }
};

3.2 Binary Tree Maximum Path Sum Java

class Solution {
    int max = Integer.MIN_VALUE;
    public int maxPath(TreeNode root) {
        if(root == null) return 0;
        int value = root.val;
        int left_sum = Math.max(maxPath(root.left),0);
        int right_sum = Math.max(maxPath(root.right),0);
        max = Math.max(max, left_sum + right_sum + value);
        return Math.max(left_sum, right_sum) + value;
    }
    public int maxPathSum(TreeNode root) {
        maxPath(root);
        return max;
    }
}

3.3 Binary Tree Maximum Path Sum JavaScript

var maxPathSum = function (root) {
  const ans = { val: -Infinity };
  dfs(root, ans);
  return ans.val;
};
function dfs(root, ans) {
  if (!root) return 0;
  const left = dfs(root.left, ans);
  const right = dfs(root.right, ans);
  const maxVal = Math.max(root.val, root.val + left, root.val + right);
  ans.val = Math.max(ans.val, maxVal, root.val + left + right);
  return maxVal;
}

3.4 Binary Tree Maximum Path Sum Python

class Solution(object):
    def maxPathSum(self, root):
        ans = [root.val]
        def DFS(root):
            if root == None:
                return 0
            lmax = max(0, DFS(root.left))
            rmax = max(0, DFS(root.right))
            ans[0] = max(ans[0] , root.val + lmax + rmax)
            return root.val + max(lmax , rmax)
        DFS(root) 
        return ans[0]

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(n)	O(h)
Java	O(n)	O(h)
JavaScript	O(n)	O(h)
Python	O(n)	O(h)

where H is the height of the tree.