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
Following code uses classic BFS but push right child first; before pushing, check the back of the queue to see whether it is in the same level. If the answer is yes, set the next pointer.

This solution always work for the next question. AND BUG FREE! :) I like BFS/DFS!

However, it does not satisfy the constant space requirement.


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