107. 二叉树的层次遍历II

版权声明：随意取用(´･ω･`) / https://blog.csdn.net/square_zou/article/details/84405421

Problem

给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

例如：
给定二叉树 [3,9,20,null,null,15,7],
返回其自底向上的层次遍历为：

too young 思路

遍历然后print。。。？
尝试写了个错的出来。。

dalao思路

看的我很是痛苦_(:」∠)
dfs-recursively-28ms

# 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 levelOrderBottom(self, root):
        # with dfs recursivrly
        value = []
        self.dfs(root, 0, value)
        return value
        
    def dfs(self, root, level, value):
        if root:
            if len(value) < level + 1:
                value.insert(0, [])
                
            value[-(level+1)].append(root.val)
            self.dfs(root.left, level+1, value)
            self.dfs(root.right, level+1, value)

dfs-stack-28ms

# 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 levelOrderBottom(self, root):
        # with dfs stackd
        stack = [(root, 0)]
        value = []
        
        while stack:
            node, level = stack.pop()
            if node:
                if len(value) < level + 1:
                    value.insert(0, [])
                value[-(level+1)].append(node.val)
                
                stack.append((node.right, level+1))   #先进去的后出来
                stack.append((node.left, level+1))
        
        return value

bfs-queue-28ms

# 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 levelOrderBottom(self, root):
        # with bfs queue
        queue = collections.deque([(root, 0)])#双端队列
        value = []
        
        while queue:
            node, level = queue.popleft()
            if node:
                if len(value) < level + 1:
                    value.insert(0, [])
                value[-(level+1)].append(node.val)
                queue.append((node.left, level+1))   
                queue.append((node.right, level+1))
        
        return value