Populating Next Right Pointers in Each Node II -- LeetCode

原题链接: http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
这道题目的要求和Populating Next Right Pointers in Each Node是一样的，只是这里的二叉树没要求是完全二叉树。其实在实现Populating Next Right Pointers in Each Node的时候我们已经兼容了不是完全二叉树的情况，其实也比较好实现，就是在判断队列结点时判断一下他的左右结点是否存在就可以了。具体算法就不再分析了，不熟悉的朋友可以看看Populating Next Right Pointers in Each Node。时间复杂度和空间复杂度不变，还是O(n)和O(1)。代码如下：

public void connect(TreeLinkNode root) {    if(root == null)        return;    TreeLinkNode lastHead = root;    TreeLinkNode pre = null;    TreeLinkNode curHead = null;    while(lastHead!=null)    {        TreeLinkNode lastCur = lastHead;        while(lastCur != null)        {            if(lastCur.left!=null)            {                if(curHead == null)                {                    curHead = lastCur.left;                    pre = curHead;                }                else                {                    pre.next = lastCur.left;                    pre = pre.next;                }            }            if(lastCur.right!=null)            {                if(curHead == null)                {                    curHead = lastCur.right;                    pre = curHead;                }                else                {                    pre.next = lastCur.right;                    pre = pre.next;                }            }                            lastCur = lastCur.next;        }        lastHead = curHead;        curHead = null;    }}

这道题本质是树的层序遍历，只是队列改成用结点自带的链表结点来维护。