# 117. Populating Next Right Pointers in Each Node II

## 刷题内容

• https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii

Given a binary tree

}
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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
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


## 解题方案

class Solution:
# @param root, a tree link node
# @return nothing
def connect(self, root):
res = []
self.recurHelper(root, 0, res)
for cur_level in res:
for i in range(len(cur_level)-1):
cur_level[i].next = cur_level[i+1]

def recurHelper(self, root, level, res):
if not root: return
if len(res) < level + 1:
res.append([])
res[level].append(root)
self.recurHelper(root.left, level+1, res)
self.recurHelper(root.right, level+1, res)


class Solution:
# @param root, a tree link node
# @return nothing
def connect(self, root):
if not root:
return
res, cur_level = [], [root]
while cur_level:
next_level = []
for node in cur_level:
if node.left:
next_level.append(node.left)
if node.right:
next_level.append(node.right)
res.append(next_level)
cur_level = next_level

for cur_level in res:
for i in range(len(cur_level)-1):
cur_level[i].next = cur_level[i+1]