Populating Next Right Pointers in Each Node I or II

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

 

For example,
Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7

 

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

我以前为啥要做那么复杂呢

ref http://www.cnblogs.com/springfor/p/3889327.html

关键在于找到有效的下家P

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