LeetCode-116-Populating Next Right Pointers in Each Node

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

解题思路：用三个指针，父指针用于循环遍历。子指针用于子节点的遍历。子头指针用于记录每一层子指针的头。父指针经历两个循环，外循环用于从上往下遍历，内循环用于层内水平遍历。

    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;
        }
        
    }