LeetCode-in-Java

124. Binary Tree Maximum Path Sum

Hard

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:

To solve the “Binary Tree Maximum Path Sum” problem in Java with a Solution class, we’ll use a recursive approach. Below are the steps:

  1. Create a Solution class: Define a class named Solution to encapsulate our solution methods.

  2. Create a maxPathSum method: This method takes the root node of the binary tree as input and returns the maximum path sum.

  3. Define a recursive helper method: Define a recursive helper method maxSumPath to compute the maximum path sum rooted at the current node.
    • The method should return the maximum path sum that can be obtained from the current node to any of its descendants.
    • We’ll use a post-order traversal to traverse the tree.
    • For each node:
      • Compute the maximum path sum for the left and right subtrees recursively.
      • Update the maximum path sum by considering three cases:
        1. The current node itself.
        2. The current node plus the maximum path sum of the left subtree.
        3. The current node plus the maximum path sum of the right subtree.
      • Update the global maximum path sum if necessary by considering the sum of the current node, left subtree, and right subtree.

  4. Initialize a variable to store the maximum path sum: Initialize a global variable maxSum to store the maximum path sum.

  5. Call the helper method: Call the maxSumPath method with the root node.

  6. Return the maximum path sum: After traversing the entire tree, return the maxSum.

Here’s the Java implementation:

class Solution {
    int maxSum = Integer.MIN_VALUE; // Initialize global variable to store maximum path sum
    
    public int maxPathSum(TreeNode root) {
        maxSumPath(root);
        return maxSum; // Return maximum path sum
    }
    
    // Recursive helper method to compute maximum path sum rooted at current node
    private int maxSumPath(TreeNode node) {
        if (node == null) return 0; // Base case
        
        // Compute maximum path sum for left and right subtrees recursively
        int leftSum = Math.max(maxSumPath(node.left), 0); // Ignore negative sums
        int rightSum = Math.max(maxSumPath(node.right), 0); // Ignore negative sums
        
        // Update maximum path sum by considering three cases:
        // 1. Current node itself
        // 2. Current node + maximum path sum of left subtree
        // 3. Current node + maximum path sum of right subtree
        int currentSum = node.val + leftSum + rightSum;
        maxSum = Math.max(maxSum, currentSum); // Update global maximum path sum
        
        // Return the maximum path sum that can be obtained from the current node to any of its descendants
        return node.val + Math.max(leftSum, rightSum);
    }
    
    // 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;
        }
    }
}

This implementation follows the steps outlined above and efficiently computes the maximum path sum in a binary tree in Java.

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