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.
解题思路：
DFS暴力枚举，注意，如果采用static 全局变量的话，在IDE里面是可以通过，但在OJ上无法测试通过，因此需要建立一个类来储存结果，JAVA实现如下：
public class Solution {
static public int maxPathSum(TreeNode root) {
Result result=new Result(Integer.MIN_VALUE);
dfs(root,result);
return result.val;
}
static int dfs(TreeNode root,Result result) {
if (root == null)
return 0;
int left_sum = Math.max(0, dfs(root.left,result));
int right_sum = Math.max(0, dfs(root.right,result));
result.val = Math.max(result.val, left_sum + right_sum + root.val);
return Math.max(left_sum, right_sum) + root.val;
}
}
class Result{
int val;
Result(){
this.val=Integer.MIN_VALUE;
}
Result(int val){
this.val=val;
}
}
Java for LeetCode 124 Binary Tree Maximum Path Sum
原文：http://www.cnblogs.com/tonyluis/p/4531152.html