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.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

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

思路：题目明确要求不许用额外空间，递归栈除外。所以这一道标准的二叉树遍历。

处理完左右节点后，要将左右子树链接起来，而不仅仅是左右节点，这点切记！！如果只连接左右节点的话，当处理层数为4的完全二叉树时，会第4层的前4个节点与后4个节点是分离的！！

写法1：

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public TreeLinkNode dfs(TreeLinkNode node){
        if(node == null) return null;
        if(node.left == null && node.right == null) return node;
        TreeLinkNode lnode = dfs(node.left);
        TreeLinkNode rnode = dfs(node.right);
        lnode.next = rnode;
        if(lnode.right != null && rnode.left != null){
            TreeLinkNode lrSonNode= lnode.right;
            TreeLinkNode rlSonNode= rnode.left;
            while(lrSonNode != null &&rlSonNode != null){
                lrSonNode.next = rlSonNode;
                lrSonNode = lrSonNode.right;
                rlSonNode = rlSonNode.left;
            }
        }
        return node;
    }
    
    public void connect(TreeLinkNode root) {
        dfs(root);
    }
}

写法2：

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        if( root == null )
            return;
        if( root.left != null )
        {
            root.left.next = root.right;
        }
        
        if( root.right != null)
        {
            if( root.next != null )
            {
                root.right.next = root.next.left;
            }
        }
        
        connect( root.left );
        connect( root.right );
    }
    
    
}