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  • 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。

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  • 原文地址:https://www.cnblogs.com/sherylwang/p/5483685.html
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