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);
}
}