hard Path with Maximum Sum
Problem Statement
Find the path with the maximum sum in a given binary tree. Write a function that returns the maximum sum.
A path can be defined as a sequence of nodes between any two nodes and doesn’t necessarily pass through the root. The path must contain at least one node.
Constraints:
- The number of nodes in the tree is in the range [1, 3 * 104].
-1000 <= Node.val <= 1000
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✅ Solution Path with Maximum Sum
Problem Statement
Find the path with the maximum sum in a given binary tree. Write a function that returns the maximum sum.
A path can be defined as a sequence of nodes between any two nodes and doesn’t necessarily pass through the root. The path must contain at least one node.
Constraints:
- The number of nodes in the tree is in the range [1, 3 * 104].
-1000 <= Node.val <= 1000
Solution
This problem follows the Binary Tree Path Sum pattern and shares the algorithmic logic with Tree Diameter. We can follow the same DFS approach. The only difference will be to ignore the paths with negative sums. Since we need to find the overall maximum sum, we should ignore any path which has an overall negative sum.
Here is the visual representation of the algorithm:
Code
Here is the code for this algorithm:
// class TreeNode {
// int val;
// TreeNode left;
// TreeNode right;
// TreeNode(int x) {
// val = x;
// }
// };
class Solution {
private static int globalMaximumSum;
public int findMaximumPathSum(TreeNode root) {
globalMaximumSum = Integer.MIN_VALUE;
findMaximumPathSumRecursive(root);
return globalMaximumSum;
}
private static int findMaximumPathSumRecursive(TreeNode currentNode) {
if (currentNode == null) return 0;
int maxPathSumFromLeft = findMaximumPathSumRecursive(currentNode.left);
int maxPathSumFromRight = findMaximumPathSumRecursive(currentNode.right);
// ignore paths with negative sums, since we need to find the maximum sum we should
// ignore any path which has an overall negative sum.
maxPathSumFromLeft = Math.max(maxPathSumFromLeft, 0);
maxPathSumFromRight = Math.max(maxPathSumFromRight, 0);
// maximum path sum at the current node will be equal to the sum from the left
// subtree + the sum from right subtree + val of current node
int localMaximumSum =
maxPathSumFromLeft + maxPathSumFromRight + currentNode.val;
// update the global maximum sum
globalMaximumSum = Math.max(globalMaximumSum, localMaximumSum);
// maximum sum of any path from the current node will be equal to the maximum of
// the sums from left or right subtrees plus the value of the current node
return Math.max(maxPathSumFromLeft, maxPathSumFromRight) + currentNode.val;
}
public static void main(String[] args) {
Solution sol = new Solution();
TreeNode root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
System.out.println("Maximum Path Sum: " + sol.findMaximumPathSum(root));
root.left.left = new TreeNode(1);
root.left.right = new TreeNode(3);
root.right.left = new TreeNode(5);
root.right.right = new TreeNode(6);
root.right.left.left = new TreeNode(7);
root.right.left.right = new TreeNode(8);
root.right.right.left = new TreeNode(9);
System.out.println("Maximum Path Sum: " + sol.findMaximumPathSum(root));
root = new TreeNode(-1);
root.left = new TreeNode(-3);
System.out.println("Maximum Path Sum: " + sol.findMaximumPathSum(root));
}
}
Time Complexity
The time complexity of the above algorithm is
Space Complexity
The space complexity of the above algorithm will be
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