【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

Return6.

思路：

左子树延伸下去的路径

右子树眼神下去的路径

通过根节点的路径

注意返回这和最大值的关系；

class Solution{int help(TreeNode* node, int& maxVal){if (!node){return 0;}int v1 = help(node->left, maxVal);int v2 = help(node->right, maxVal);int v = max(max(v1 + node->val, v2 + node->val), max(v1 + v2 + node->val, node->val));maxVal = v>maxVal ? v : maxVal;return max(max(v1 + node->val, v2 + node->val), node->val);//返回链接父节点时的最大值}int maxPathSum(TreeNode* root){int maxVal = INI_MIN;help(root, maxVal);return maxVal;}};