rn nullptr; } // The root's start and end is given by build method. SegmentTreeNode *root = new SegmentTreeNode(start, end, INT_MAX); // If start equals to end, there will be no children for this node. if (start == end) { root->max = A[start]; return root; } // Left child: start=A.left, end=(A.left + A.right) / 2. root->left = buildRecu(A, start, (start + end) / 2); // Right child: start=(A.left + A.right) / 2 + 1, end=A.right. root->right = buildRecu(A, (start + end) / 2 + 1, end); const int left_max = root->left != nullptr ? root->left->max : INT_MAX; const int right_max = root->right != nullptr ? root->right->max : INT_MAX; // Update max. root->max = max(left_max, right_max); return root; } }; 4:["$","div",null,{"children":[["$","header",null,{"children":[["$","div",null,{"className":"text-3xl text-bold p-0 mt-2 mb-2","children":["$","h1",null,{"children":["Segment Tree Build II",["$","span",null,{"className":"font-mono text-sm text-gray-500","children":"(LintCode 439):"}]]}]}],""]}],["$","article",null,{"children":[["$","div",null,{"className":"p-0","dangerouslySetInnerHTML":{"__html":"

Problem description: Visit and Support LintCode Search via
Difficulty: Medium
Tags: LintCode, Recursion, O(n), O(h), Medium, Segmen

Problem