题目描述:
给定一个非空二叉树, 返回一个由每层节点平均值组成的数组.
示例 1:
输入:
3
/ \
9 20
/ \
15 7
输出: [3, 14.5, 11]
解释:
第0层的平均值是 3, 第1层是 14.5, 第2层是 11. 因此返回 [3, 14.5, 11].
解题思路:
二叉树的层次遍历,每一层求平均值。
代码实现:
# 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 averageOfLevels(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if(root == None):
return []
else:
res = []
queue = [root]
while(len(queue) > 0):
tempList = []
for i in range(len(queue)):#里层for循环是对于每一个节点,都判断其左右孩子
temp = queue.pop(0)
tempList.append(temp.val)
if(temp.left):
queue.append(temp.left)
if(temp.right):
queue.append(temp.right)
res.append(sum(tempList)/len(tempList))
# result = []
# for i in res:
# temp = float(sum(i)/len(i))
# result.append(temp)
return res
n1 = TreeNode(4)
n2 = TreeNode(2)
n3 = TreeNode(7)
n4 = TreeNode(1)
n5 = TreeNode(3)
n1.left = n2
n1.right = n3
n2.left = n4
n2.right = n5
s = Solution()
result = s.averageOfLevels(n1)
print(result)