117. Populating Next Right Pointers in Each Node II -----M

Given a binary tree
struct Node {
  int val;
  Node *left;
  Node *right;
  Node *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.
 
Follow up:
You may only use constant extra space.
Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
 
Example 1:
Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
 
Constraints:
The number of nodes in the given tree is less than 6000.
-100 <= node.val <= 100

A:

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        Node * preRunner = root; // initially put at the left most
        while(preRunner){
            Node * curLeftMost = nullptr;
            Node * curLevelRunner = nullptr;
            while(preRunner){
                if(preRunner->left ){
                    if(!curLeftMost){
                        curLeftMost = preRunner->left;
                        curLevelRunner = curLeftMost;
                    }else{
                        curLevelRunner->next = preRunner->left;
                        curLevelRunner =  curLevelRunner->next;
                    }
                }
                if(preRunner->right ){
                    if(!curLeftMost){
                        curLeftMost = preRunner->right;
                        curLevelRunner = curLeftMost;
                    }else{
                        curLevelRunner->next = preRunner->right;
                        curLevelRunner =  curLevelRunner->next;
                    }
                }
                preRunner = preRunner->next;
            }
            preRunner = curLeftMost;
        }
        return root;
    }
};

Mistakes:
1: upperRunner 并不需要每次都update，如果lowerRunner是left child,并且，他还有right child,就不必。（看上面代码）


Learned:

-------------------第二遍，没有像第一遍那样，extract出几个函数来-----------好多层while重叠， 搞得很乱------------

public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode pHeader = new TreeLinkNode(0);
        pHeader.next = root;
        while(pHeader.next !=null){
            TreeLinkNode header = new TreeLinkNode(0);
            TreeLinkNode pRunner = pHeader.next;
            TreeLinkNode runner = header;
            while(pRunner!=null){
                if(pRunner.left != null){
                    runner.next = pRunner.left;
                    runner = runner.next;
                }
                if(pRunner.right != null ){
                    runner.next = pRunner.right;
                    runner = runner.next;
                }
                pRunner = pRunner.next;
            }
            pHeader = header;        
        }    
    }
}