Populating Next Right Pointers in Each Node II

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

这道题延续Populating Next Right Pointers in Each Node 的思路,使用层搜和上一层的linkedlist连接实现O(1)的空间复杂度，但是实际实现时因为不是完全二叉树，不能直接通过上一层建立下一层的连接(时间复杂度高）。而需要同时保存下一层的prev已方便连接。实现时需要判断当前上层结点是否有左右子树，代码如下：

class Solution(object):def connect(self, root):""":type root: TreeLinkNode:rtype: nothing"""if not root:returnhead = root    #low level's start nodeprev = None    #low level's previous nodeupCur = None   # higher level's currrent nodewhile(head):upCur = head prev = Nonehead = Nonewhile upCur:if upCur.left:if prev == None:prev = upCur.lefthead = prevelse:prev.next = upCur.leftprev = prev.next if upCur.right:if prev == None:prev = upCur.righthead = prevelse:prev.next = upCur.rightprev = prev.nextupCur = upCur.next

另外一种需要判断的代码如下：

class Solution(object):def connect(self, root):""":type root: TreeLinkNode:rtype: nothing"""if not root:returnleftMost = root               #higher level point while leftMost:prev = leftMostwhile prev and (not prev.left) and (not prev.right):  #no low level nodes should be connectedprev = prev.nextif not prev:              #the lowest level of treereturn leftMost = prev.left if prev.left else prev.right  #for the next loop, leftMost is not Nonecur = leftMostwhile(prev):              #higer level's node if cur == prev.left:if prev.right:    #if root.right is not None, find the child of other nodescur.next = prev.rightcur = cur.nextprev = prev.nextelif cur == prev.right:  prev = prev.next  #can't connected directedly.else:if (not prev.left) and (not prev.right):prev = prev.nextcontinue      #can't find cur.next through this rootcur.next = prev.left if prev.left else prev.rightcur = cur.next

这种解法分情况讨论目前遍历到的下一层结点和目前当前层结的关系，讨论比较复杂。不如上面那种解法方便。

转载于:https://www.cnblogs.com/sherylwang/p/5483685.html

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