LintCode二叉树中的最大路径和

给出一棵二叉树，寻找一条路径使其路径和最大，路径可以在任一节点中开始和结束（路径和为两个节点之间所在路径上的节点权值之和）

样例
给出一棵二叉树：

   1
  / \
 2   3

返回 6

代码:

/**
 * 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.
     */
   int max = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        if(null == root)
        {
            return 0;
        }
        subMaxPathSum(root);
        return max;
        }
    
    public int subMaxPathSum(TreeNode node)
    {
        if(null == node)
        {
            return 0;
        }
        int leftMax = subMaxPathSum(node.left);
        int rightMax = subMaxPathSum(node.right);
        int subMax = Math.max(Math.max(Math.max(leftMax + node.val, rightMax + node.val), node.val),leftMax + rightMax + node.val);
        if(max < subMax)
        {
            max = subMax;
        }
        return Math.max(Math.max(leftMax + node.val, rightMax + node.val), node.val);
    }
}