# LEETCODE —— Populating Next Right Pointers in Each Node

### Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {
}


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

while(cursor_high.left!=None):

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



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