Populating Next Right Pointers in Each Node II -- LeetCode
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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:
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。
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