LEETCODE —— Populating Next Right Pointers in Each Node

Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1
       /  
      2    3
     /   / 
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  
      2 -> 3 -> NULL
     /   / 
    4->5->6->7 -> NULL

'''
Created on Nov 19, 2014

@author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com>
'''
# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        stack=[]
        if(root==None): return
        stack.append(root)
        while(len(stack)!=0):
            for ix in range(len(stack)-1):
                stack[ix].next=stack[ix+1]
            stack[-1].next=None
            cntOfLastLevel=len(stack)
            for ix in range(cntOfLastLevel):
                if (stack[0].left!=None):stack.append(stack[0].left)
                if (stack[0].right!=None):stack.append(stack[0].right)
                stack.remove(stack[0])
        
class Solution2:
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        if(root==None): return
        head_high=cursor_high=root
        head_low = cursor_low=None
        
        while(cursor_high.left!=None):
            head_low = cursor_low=cursor_high.left
         
            cursor_low.next=cursor_high.right
            cursor_low=cursor_low.next
            while(cursor_high.next!=None):
                cursor_high=cursor_high.next
                cursor_low.next=cursor_high.left
                cursor_low=cursor_low.next
                cursor_low.next=cursor_high.right
                cursor_low=cursor_low.next
            cursor_low.next=None
              
            head_high=cursor_high=head_low
        

题目和代码如上所述，这个题比较有意思的地方是，这个数据结构有点像数据库索引的结构，上述代码实现了两种方法，两种方法都是层级遍历，时间复杂度都是O(N)，但空间复杂度不一样，实现难度也不一样。

1. 第一种更为简单，但使用了额外的空间用于存储上一层节点，最大空间复杂度为所有叶子节点的大小之和。所以类似这种算法如果用于建立DB索引的话，恐怕内存就要爆掉了，第二种方法则没有问题。

2. 第二种稍复杂，但是空间复杂度只有O(1)，也就是无需任何额外内存。实现方法是使用两个游标和1个标志位，两个游标用于并行遍历两行nodes，1个标志位用于标记下面那行的head。

　两游标并行往前走，会同时走到各自行的结尾，这时两游标各自下移到下一行开始（这就是为啥要标记那行的head）然后重复上面的过程继续往前走，当没有下一行时停止（第二个游标没得指了），请自行脑补2个游标遍历所有nodes并且加链接的过程。