LeetCode(124) Binary Tree Maximum Path Sum

题目如下：

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.

分析如下：

注意从NODE A开始向叶节点方向增长的的maximum path一共有一下这4种情况

1 NODE, 如下面这种情况，最大值就是node(value = 1)自己。

       1
      / \
     -2   -3

2 NODE  + left child

       1
      / \
     2   -3

3 NODE  + right child

       1
      / \
     -2   3

4 NODE  + right child + right child

       1
      / \
     2   3

其中除了4是折线(下面代码中的twofold_path_max)外，1~3都是直线(下面代码中额left_onefold_max 和right_onefold_max)。

对于每个节点算出的这四个值，都和保存当前最大值的变量maximum_result进行比较，不断地刷新maximum_result的值。

我的代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int my_maxPathSum(TreeNode *root, int onefold_path_max, int twofold_path_max, int& max_result) {        if (root == NULL) return 0;        int left_onefold_max = my_maxPathSum(root->left, onefold_path_max, twofold_path_max, max_result);         int right_onefold_max = my_maxPathSum(root->right, onefold_path_max, twofold_path_max, max_result);        int return_one_fold_max = 0;        twofold_path_max = left_onefold_max + right_onefold_max + root->val; // node + left + right，作为最大值的候选        left_onefold_max = (left_onefold_max)>0?(left_onefold_max + root->val):(root->val); //从node和node + left中选个最大值        right_onefold_max = (right_onefold_max)>0?(right_onefold_max + root->val):(root->val);//从node和node + right中选个最大值        return_one_fold_max = (right_onefold_max > left_onefold_max)? right_onefold_max:left_onefold_max;        max_result = (max_result > right_onefold_max)? max_result:right_onefold_max; //Note: max_result是轮流地和candidate比较        max_result = (max_result > left_onefold_max)? max_result:left_onefold_max;        max_result = (max_result > twofold_path_max)? max_result:twofold_path_max;        return return_one_fold_max;    }    int maxPathSum(TreeNode *root) {        int max_result = INT_MIN;        int onefold_path_max = 0;        int twofold_path_max = 0;        my_maxPathSum(root, onefold_path_max, twofold_path_max, max_result);        return max_result;    }};