Binary Tree Maximum Path Sum Algorithm

Question

I am new to recursion and binary trees. I am trying to solve this problem on leetcode.

To find maximum sum path is like finding maximum path between any two nodes, that path may or may not pass through the root; except that with max sum path we want to track sum instead of path length.

My algorithm is passing 91/93 test cases and I just can't figure out what I am missing. Can anyone please provide some direction?

class Solution {
    private int sum = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        maxPathSumHelper(root);
        if(root.left != null){
            maxPathSumHelper(root.left);
        }
        if(root.right != null){
            maxPathSumHelper(root.right);
        }
        
        return sum;
    }
    public int  maxPathSumHelper(TreeNode root){
        if(root ==  null){
            return 0;
        }
        //check left sum
        int leftValue = root.val + maxPathSumHelper(root.left);
        if(leftValue > sum){
            sum = leftValue;
        }
        //check right sum
        int rightValue = root.val + maxPathSumHelper(root.right);
        if(rightValue > sum){
            sum = rightValue;
        }
        //check if root value is greater 
        if(root.val > sum){
            sum = root.val;
        }
        //check if right and left value is the greatest
        if((leftValue + rightValue - (2 * root.val) )+ root.val > sum){
            sum = (leftValue + rightValue - (2 * root.val)) + root.val;
        }
        return Math.max(leftValue, rightValue);
    }
}

Answer 1

Try

class Solution {
    private int sum = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        maxPathSumHelper(root);

        if (root.left != null) {
            maxPathSumHelper(root.left);

        } else if (root.right != null) {
            maxPathSumHelper(root.right);

        }

        return sum;
    }
    public int  maxPathSumHelper(TreeNode root) {
        if (root ==  null) {
            return 0;
        }

        //check left sum
        int leftValue = root.val + maxPathSumHelper(root.left);

        if (leftValue > sum) {
            sum = leftValue;
        }

        //check right sum
        int rightValue = root.val + maxPathSumHelper(root.right);

        if (rightValue > sum) {
            sum = rightValue;
        }

        //check if root value is greater
        if (root.val > sum) {
            sum = root.val;
        }

        //check if right and left value is the greatest
        if ((leftValue + rightValue - (2 * root.val) ) + root.val > sum) {
            sum = (leftValue + rightValue - (2 * root.val)) + root.val;
        }

        return Math.max(Math.max(leftValue, rightValue), root.val);
    }
}

Answer 2

I guess we can a bit simplify our statements here.

This'll simply get accepted:

public final class Solution {
    int max;

    public final int maxPathSum(final TreeNode root) {
        max = Integer.MIN_VALUE;
        traverse(root);
        return max;
    }

    private final int traverse(final TreeNode node) {
        if (node == null)
            return 0;
        final int l = Math.max(0, traverse(node.left));
        final int r = Math.max(0, traverse(node.right));
        max = Math.max(max, l + r + node.val);
        return Math.max(l, r) + node.val;
    }
}

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References

• For additional details, please see the Discussion Board which you can find plenty of well-explained accepted solutions in there, with a variety of languages including efficient algorithms and asymptotic time / space complexity analysis ^{1 , 2} .