算法描述：

Given a binary tree

struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

Example:

Given the following perfect binary tree,

     1
   /    2    3
 / \  / 4  5  6  7

After calling your function, the tree should look like:

     1 -> NULL
   /    2 -> 3 -> NULL
 / \  / 4->5->6->7 -> NULL

解題思路：用三個指針，父指針用於循環遍歷。子指針用於子節點的遍歷。子頭指針用於記錄每一層子指針的頭。父指針經歷兩個循環，外循環用於從上往下遍歷，內循環用於層內水平遍歷。

    void connect(TreeLinkNode *root) {
        TreeLinkNode
 
* childHead = nullptr;
        TreeLinkNode* child = nullptr;
        while(root!=nullptr){
            while(root!=nullptr){
                if(root->left!=nullptr){
                    if(childHead!=nullptr)
                        child->next = root->left;
                    else 
                        childHead = root->left;
                    
                    child = root->left;
                }
                
                if(root->right != nullptr){
                    if(childHead!=nullptr)
                        child->next = root->right;
                    else
                        childHead = root->right;
                    child = root->right;
                }
                
                root = root->next;
            }
            root = childHead;
            childHead = nullptr;
            child = nullptr;
        }
        
    }

LeetCode-116-Populating Next Right Pointers in Each Node