【leetcode】 Populating Next Right Pointers in Each Node

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 toNULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For 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

这题的精典在于从右往左的处理！！如果从左往右，就会出现漏掉的情况！！
如果发现对于single linklist，需要去access前一个节点时，不如换个角度，从当前点出发去处理后一个节点。

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL)
            return;
         connet(root, root->right);    
        connet(root, root->left);
       
        
    }
    
    void connet(TreeLinkNode* parent, TreeLinkNode* current){
        if (current == NULL)
            return; 
            
        if (current == parent->left && parent->right != NULL)
            current->next = parent->right;
            
        else{
            TreeLinkNode* parentNext = parent->next;
            if (parentNext!=NULL){
                TreeLinkNode* currentNext = NULL;
                while(parentNext &¤tNext==NULL ){
                    currentNext = parentNext->left;
                    if (currentNext==NULL)
                        currentNext = parentNext->right;
                    parentNext = parentNext->next;
                }
                current->next = currentNext;
                
            }
        }
        
        connet(current, current->right);
        connet(current, current->left);
    }
};