LeetCode in Python 102. Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

 

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

Solution:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        q, res = root and [root], []
        while q:
            size, level_path = len(q), []
            for i in range(size):
                node = q.pop(0)
                level_path.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            res.append(level_path)
        return res
        

最经典的队列实现BFS，注意size是作为for循环次数的依据，Python的 for node in q 的循环不适用，因为q在动态变化。

N-ary的树可参考559（https://www.cnblogs.com/lowkeysingsing/p/11152751.html）