g0101_0200.s0124_binary_tree_maximum_path_sum.Solution Maven / Gradle / Ivy

Go to download
package g0101_0200.s0124_binary_tree_maximum_path_sum;

// #Hard #Top_100_Liked_Questions #Top_Interview_Questions #Dynamic_Programming #Depth_First_Search
// #Tree #Binary_Tree #Udemy_Tree_Stack_Queue #Big_O_Time_O(N)_Space_O(N)
// #2022_06_23_Time_1_ms_(99.46%)_Space_47.2_MB_(77.68%)

import com_github_leetcode.TreeNode;

/*
 * 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;
 *     }
 * }
 */
/**
 * 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/13/exx1.jpg)
 *
 * **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:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/13/exx2.jpg)
 *
 * **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`
**/
public class Solution {
    private int max = Integer.MIN_VALUE;

    private int helper(TreeNode root) {
        if (root == null) {
            return 0;
        }
        // to avoid the -ve values in left side we will compare them with 0
        int left = Math.max(0, helper(root.left));
        int right = Math.max(0, helper(root.right));
        int current = root.val + left + right;
        if (current > max) {
            max = current;
        }
        return root.val + Math.max(left, right);
    }

    public int maxPathSum(TreeNode root) {
        helper(root);
        return max;
    }
}