116. Populating Next Right Pointers in Each Node Leetcode Java

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

Solution:

It is similar to the level order traversal, but here we do not need an extra queue to get he next unprocessed node, we can use the next pointer to do that. Here I maintain a firstNode to store the firstNode of each level, so when we finished processing the current level, we can start to process next level by iterate to firstNode.left. We can do that because it is perfect BT. For any BT, please refer to next problem: 117. Populating Next Right Pointers in Each Node II. 

  public void connect(TreeLinkNode root) {  
     if(root==null) return;  
     TreeLinkNode cur=root;  
     TreeLinkNode firstNode=root;  
     while(cur!=null){  
       if(cur.left!=null && cur.right!=null){  
         cur.left.next=cur.right;  
        if(cur.next!=null) {  
         cur.right.next=cur.next.left;  
         cur=cur.next;   
        }  
        else{  
         cur=firstNode.left;  
         firstNode=cur;  
       }  
       }  
       else cur=cur.next;  
     }  
   }