题目：

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

思路：

递归，对root的左右进行连接操作，然后先递归右树，再递归左树。

package tree;public class PopulatingNextRightPointersInEachNodeII {    public void connect(TreeLinkNode root) {        if (root == null || (root.left == null && root.right == null)) return;        TreeLinkNode next = root.next;        while (next != null) {            if(next.left != null) {                next = next.left;                break;            }                        if (next.right != null) {                next = next.right;                break;            }                        next = next.next;        }                if (root.right != null)             root.right.next = next;                 if (root.left != null)            root.left.next = root.right == null ? next : root.right;                        connect(root.right);        connect(root.left);    }}