Binary Tree Maximum Path Sum [leetcode] dp

a(i)：以节点i作为终点的单边最大路径和

b(i)：以节点i作为终点的双边边最大路径和

a(i) = max{ i->val,i->val + max{a(i->left), a(i->right) }};

b(i) = max{ i->val, i->val + max{a(i->left), a(i->right) } ,i->val + a(i->left) + a(i->right)};

由于a(i), b(i)仅仅和a(i->left)和a(i->right) 有关，因此可以将空间压缩为O(1)

代码如下：

    int maxPathSum(TreeNode *root) {        int res = INT_MIN;        getSum(root, res);        return res;    }        int getSum(TreeNode * root, int & res)    {        if (root == NULL) return 0;        int l = getSum(root->left, res);        int r = getSum(root->right, res);        int a, b;        a = max(root->val, root->val + max(l, r));//one side        b = max(a, root->val + l + r);            //both side        res = max(res, max(a, b));        return a;    }