[lintcode] Binary Tree Maximum Path Sum II

Given a binary tree, find the maximum path sum from root.

The path may end at any node in the tree and contain at least one node in it.

给一棵二叉树，找出从根节点出发的路径中，和最大的一条。

这条路径可以在任何二叉树中的节点结束，但是必须包含至少一个点（也就是根了）。

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root the root of binary tree.
     * @return an integer
     */
    public int maxPathSum2(TreeNode root) {
        if (root ==  null) {
            return Integer.MIN_VALUE;
        }
 
        int left = maxPathSum2(root.left);
        int right = maxPathSum2(root.right);
 
        return root.val + Math.max(0, Math.max(left, right));
    }
}