Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
1 / \ 2 3
Return 6
.
/** * Definition for binary tree public class TreeNode { int val; TreeNode left; * TreeNode right; TreeNode(int x) { val = x; } } */ public class Solution { public int maxPathSum(TreeNode root) { // Start typing your Java solution below // DO NOT write main() function Max max = new Max(); max.num = root.val; maxPathSum(root, max); return max.num; } public int maxPathSum(TreeNode node, Max max) { if (node == null) return -1; int currentMax = node.val; int maxLeft = maxPathSum(node.left, max); int maxRight = maxPathSum(node.right, max); if (maxLeft > 0) currentMax += maxLeft; if (maxRight > 0) currentMax += maxRight; if (currentMax > max.num) max.num = currentMax; if (maxLeft > 0) { if (maxLeft > maxRight) { return node.val + maxLeft; } else { return node.val + maxRight; } } else if (maxRight > 0) { return node.val + maxRight; } else { return node.val; } } private class Max { int num; } }
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