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.

思路：

递归调用函数。用函数comput计算从子树到根节点传递到根节点的父节点的值，然后用一个result记录在递归过程中求得的最大值。

代码：

    int max(int a, int b){
        if(a > b)
            return a;
        return b;
    }
    int comput(TreeNode *root, int &result){
        if(root == NULL)
            return 0;
        int left = comput(root->left, result);
        int right = comput(root->right, result);
        int arch = left + right + root->val;
        int valToParent = max(root->val, root->val+max(left, right));
        result = max(result, max(arch, valToParent));
        return valToParent;
    }
    int maxPathSum(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int result = INT_MIN;
        comput(root, result);
        return result;
    }