[leetcode]Populating Next Right Pointers in Each Node

这道题目如果不用常数级的空间是很容易的，就是用一个queue呗。

public class Solution {

    public void connect(TreeLinkNode root) {

        LinkedList<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();

        queue.offer(root);

        int count = 0;

        int lastCount = 0;

        TreeLinkNode last = null;

        while (queue.size() != 0)

        {

            TreeLinkNode n = queue.poll();

            if (n == null) continue;

            count++;

            if (last != null) last.next = n;

            if (lastCount == 0 || count == lastCount * 2)

            {

                lastCount = count;

                count = 0;

                n.next = null;

                last = null;

            }

            else {

                last = n;

            }

            queue.offer(n.left);

            queue.offer(n.right);

        }

    }

}

但是题目要求常数级空间，怎么办呢？当时想啊，如果没有queue可怎么做到同一层从左到右遍历啊，后来搜索了之后发现其实可以用next指针嘛。其他还有递归法，精神就一样了。参见：http://blog.csdn.net/ithomer/article/details/8796542

public class Solution {

    public void connect(TreeLinkNode root) {

        TreeLinkNode left = root;

        TreeLinkNode n = null;

        while (left != null && left.left != null && left.right != null)

        {

            n = left;

            while (n != null)

            {

                n.left.next = n.right;

                if (n.next != null)

                {

                    n.right.next = n.next.left;

                }

                n = n.next;

            }

            left = left.left;

        }

    }

}