[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));
}
}