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;
    }

Binary Tree Maximum Path Sum [leetcode] dp

原文：http://blog.csdn.net/peerlessbloom/article/details/39935891