LeetCode - Populating Next Right Pointers in Each Node II

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);
}

}