Problem Link:
http://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/
Just traverse the tree from the root, level by level. Esay enough.
# 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):
"""
We use a BFS from the root, so that we can traverse the tree level by level.
We need two queues, to store the nodes of this level and next.
"""
if root is None:
return
q = [root]
while q:
new_q = []
for i in xrange(len(q)-1):
q[i].next = q[i+1]
q[-1].next = None
for node in q:
if node.left:
new_q.append(node.left)
if node.right:
new_q.append(node.right)
q = new_q