Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
解法
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
_result=collections.defaultdict(list)
#广度遍历 BFS
def _bfs(root,level,_result):
queue = [[root,1]]
while len(queue) > 0:
temp,level = queue.pop(0)
_result[level].append(temp.val)
if temp.left:
queue.append([temp.left,level+1])
if temp.right:
queue.append([temp.right,level+1])
_bfs(root,1,_result)
result=[]
for x in sorted(_result.keys(),reverse=True):
result.append(_result[x])
return result
上面的代码主要是通过队列实现BFS的访问，使用defaultdict对象保存结果，字典的key是对应每一层的编号，value就是每一层从左到右的排序结果。
下面给出的leetcode讨论区给出的答案，这个完全会更加易懂，实现思想都是基于BFS
class Solution(object):
def levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
res = []
arr = []
if root is None:
return []
else:
arr.append(root)
while True:
tmp = []
while len(arr) > 0:
item = arr.pop(0)
tmp.append(item)
res.insert(0, [w.val for w in tmp])
for item in tmp:
if item.left is not None:
arr.append(item.left)
if item.right is not None:
arr.append(item.right)
if len(arr) == 0:
break
return res
Deeplearn, 版权所有丨如未注明 , 均为原创丨本网站采用BY-NC-SA协议进行授权 , 转载请注明107. Binary Tree Level Order Traversal II(二叉树分层排序)！
