[Leetcode] Binary Tree Maximum Path Sum

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:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int result;
    public int maxPathSum(TreeNode root) {
        result=Integer.MIN_VALUE;
        dfs(root);
        return result;
    }
    private int dfs(TreeNode root) {
        // TODO Auto-generated method stub
        if(root==null)
            return 0;
        int l=0;
        int r=0;
        if(root.left!=null)
            l=dfs(root.left);
        if(root.right!=null)
            r=dfs(root.right);
        int sum=root.val;
        if(l>0)
            sum+=l;
        if(r>0)
            sum+=r;
        result=Math.max(result,sum);
        return Math.max(0,Math.max(l,r))+root.val; //只能选择左边儿的或者右边儿的来组成一个path
    }
}