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:
            return
        head = root    #low level's start node
        prev = None    #low level's previous node
        upCur = None   # higher level's currrent node
        
        while(head):
            upCur = head 
            prev = None
            head = None
            while upCur:
                if upCur.left:
                    if prev == None:
                        prev = upCur.left
                        head = prev
                    else:
                        prev.next = upCur.left
                        prev = prev.next 
                    
                if upCur.right:
                    if prev == None:
                        prev = upCur.right
                        head = prev
                    else:
                        prev.next = upCur.right
                        prev = prev.next
                upCur = 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:
            return
        leftMost = root               #higher level point 
        
        while leftMost:
            prev = leftMost
            while prev and (not prev.left) and (not prev.right):  #no low level nodes should be connected
                prev = prev.next
            if not prev:              #the lowest level of tree
                return 
            leftMost = prev.left if prev.left else prev.right  #for the next loop, leftMost is not None
            cur = leftMost
            while(prev):              #higer level's node 
                if cur == prev.left:
                    if prev.right:    #if root.right is not None, find the child of other nodes
                        cur.next = prev.right
                        cur = cur.next
                    prev = prev.next
                elif cur == prev.right:  
                    prev = prev.next  #can't connected directedly.
                else:
                    if (not prev.left) and (not prev.right):
                        prev = prev.next
                        continue      #can't find cur.next through this root
                    cur.next = prev.left if prev.left else prev.right
                    cur = 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。