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 to NULL.

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

Analyse: Recursively invoke the original function. Since it‘s perfect binary tree, if the left child does not exist, it means the current node is leaf node. Consider two situations: 

A. the left child of root directly points to the right child of the root.

B. the right child of root points to the left child of the next node of the root.

Runtime: 28ms. 

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * struct TreeLinkNode {
 4  *  int val;
 5  *  TreeLinkNode *left, *right, *next;
 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12         if(!root) return;
13         if(!root->left) return; //reaches to the leaf node
14         
15         root->left->next = root->right;
16         if(root->next)
17             root->right->next = root->next->left;
18         connect(root->left);
19         connect(root->right);
20     }
21 };