给定一个二叉树,返回其按层次遍历的节点值。 (即逐层地,从左到右访问所有节点)。
例如:
给定二叉树: [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
返回其层次遍历结果:
[
[3],
[9,20],
[15,7]
]
方法1:
递归
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def _levelOlder(self,level,result,root):
if root:
if level == len(result):
result.append([])
result[level].append(root.val)
self._levelOlder(level+1,result,root.left)
self._levelOlder(level+1,result,root.right)
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
level,result = 0,list()
self._levelOlder(level,result,root)
return result