题目：见https://leetcode.com/problems/populating-next-right-pointers-in-each-node/?tab=Description
解析1：由于该题目要求使用常数空间，因此只能使用迭代法，代码如下，注意代码也能适用该问题的第二个版本https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/?tab=Description
// 迭代法，时间复杂度 O(n)，空间复杂度 O(logn) class Solution { public: void connect(TreeLinkNode* root) { TreeLinkNode* cur = root; while (cur) { TreeLinkNode dummy{-1}; TreeLinkNode* prev = &dummy; while (cur) { if (cur -> left) { prev -> next = cur -> left; prev = prev -> next; } if (cur -> right) { prev -> next = cur -> right; prev = prev -> next; } cur = cur -> next; } cur = dummy.next; } } };解析2：该问题也可以使用递归法求解，但空间需求为 O(logn)，不满足题意，代码如下：
// 递归法，时间复杂度 O(n)，空间复杂度 O(logn) class Solution { public: void connect(TreeLinkNode* root) { connect(root, nullptr); } private: void connect(TreeLinkNode* root, TreeLinkNode* sibling) { if (!root) return; else root -> next = sibling; connect(root -> left, root -> right); if (sibling) connect(root -> right, sibling -> left); else connect(root -> right, nullptr); } };