Binary Tree Maximum Path Sum

Change-Id: I18d0bd0ddbed9517302aa674bb75c18b478395b1

Author: applewjg

BinaryTreeMaximumPathSum.java

@@ -0,0 +1,43 @@
+/*
+ Author:     Andy, nkuwjg@gmail.com
+ Date:       Jan 28, 2015
+ Problem:    Binary Tree Maximum Path Sum
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/
+ Notes:
+ 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.
+
+ Solution: Recursion...
+ */
+
+/**
+ * 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) {
+        int[] res = new int[1];
+        res[0] = Integer.MIN_VALUE;
+        maxPathSumRe(root, res);
+        return res[0];
+    }
+    int maxPathSumRe(TreeNode root, int[] res) {
+        if (root == null) return 0;
+        int left = maxPathSumRe(root.left, res);
+        int right = maxPathSumRe(root.right, res);
+        res[0] = Math.max(res[0], root.val + Math.max(left, 0) + Math.max(right, 0));
+        return Math.max(root.val, root.val + Math.max(left, right));
+    }
+}