124. Binary Tree Maximum Path Sum

1. 问题描述
   Given a binary tree, find the maximum path sum.

   For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

   For example:
   Given the below binary tree,

      1  / \ 2   3

   return 6

2. 解决思路
   定义一个全局变量存储最大值的路径。利用递归，每次递归返回对应节点的最大路径值，在递归中需要更新全部最大值路径。值得注意的是，这里存在负数的情况。。。需要判断一下条件，如果左右子树都返回负数，则递归返回的是该节点的值就好。。。

   1. 代码

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int m;    int maxPathSum(TreeNode* root) {        if (!root)            return 0;        m = INT_MIN;        int l_len = helper(root->left);        int r_len = helper(root->right);        int count = root->val;        if (l_len > 0)            count += l_len;        if (r_len > 0)            count += r_len;        if (count > m)            m = count;        return m;    }    int helper(TreeNode* root) {        if (!root)            return 0;        if (!root->left && !root->right) {            if (root->val > m)                m = root->val;            return root->val;        }        int l_len = helper(root->left);        int r_len = helper(root->right);        int count = root->val;        if (l_len > 0)            count += l_len;        if (r_len > 0)            count += r_len;        if (count > m)            m = count;        if (max(l_len,r_len) >0)            return max(l_len,r_len)+root->val;        else            return root->val;    }};